AP Chemistry Flashcards: Stoichiometry

Study Stoichiometry in AP Chemistry with focused flashcards that help you recognize the idea, recall the key rule, and apply it in practice-style prompts.

AP Chemistry

Stoichiometry

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What is the first step in a stoichiometry problem?

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ANSWER

Write a balanced chemical equation. Ensures atom conservation for stoichiometric calculations.

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This deck focuses on Stoichiometry, giving you a quick way to review the definitions, rules, and examples that matter most for AP Chemistry.

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Flashcard 1: What is the first step in a stoichiometry problem?

Answer: Write a balanced chemical equation. Ensures atom conservation for stoichiometric calculations.

Flashcard 2: What is the formula to calculate the number of particles from moles?

Answer: particles=moles×6.022×1023\text{particles} = \text{moles} \times 6.022 \times 10^{23}. Multiplies moles by Avogadro's number for particle count.

Flashcard 3: How many moles of H2H_2 are needed to produce 2 moles of NH3NH_3 in N2+3H22NH3N_2 + 3H_2 \rightarrow 2NH_3?

Answer: 3 moles. 3:2 mole ratio from balanced equation coefficients.

Flashcard 4: Calculate the percent composition of oxygen in H2OH_2O.

Answer: 88.81%. Oxygen mass (1616) divided by H2OH_2O mass (1818) ×100%\times 100\%.

Flashcard 5: Find the molar mass of CO2CO_2.

Answer: 44 g/mol. Carbon (12) + 2 oxygen atoms (16 each) = 44 g/mol.

Flashcard 6: Identify the unit for molar mass.

Answer: grams per mole (g/mol). Standard unit expressing mass per mole of substance.

Flashcard 7: Which unit is used to express concentration in stoichiometry?

Answer: Molarity (M). Standard concentration unit for solution calculations.

Flashcard 8: What is the formula for percent yield?

Answer: Percent yield=actual yieldtheoretical yield×100\text{Percent yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100. Compares actual to theoretical yield as a percentage.

Flashcard 9: Calculate the percent composition of oxygen in H2OH_2O.

Answer: 88.81%. Oxygen mass (16) divided by H2OH_2O mass (18) × 100%.

Flashcard 10: Calculate the molarity of a solution with 10 moles in 5 L.

Answer: 2 M. 10 moles5 L=2\frac{10 \text{ moles}}{5 \text{ L}} = 2 M concentration.

Flashcard 11: Identify the limiting reactant in 4 moles N2N_2 and 10 moles H2H_2 for N2+3H22NH3N_2 + 3H_2 \rightarrow 2NH_3.

Answer: H2H_2. Need 12 moles H2H_2 but only have 10; H2H_2 limits.

Flashcard 12: What is the stoichiometric coefficient for H2H_2 in 2H2+O22H2O2H_2 + O_2 \rightarrow 2H_2O?

Answer:

  1. Coefficient shows 2 molecules react in balanced equation.

Flashcard 13: What is the molar mass of H2SO4H_2SO_4?

Answer: 98.1 g/mol. 2H (2) + S (32) + 4O (64) = 98 g/mol total.

Flashcard 14: Which unit is used to express concentration in stoichiometry?

Answer: Molarity (M). Standard concentration unit for solution calculations.

Flashcard 15: What is the molecular formula for a compound with empirical formula CH2OCH_2O and molar mass 180 g/mol?

Answer: C6H12O6C_6H_{12}O_6. 180÷30=6180 \div 30 = 6, so multiply CH2OCH_2O by 6.

Flashcard 16: What is the molecular formula for a compound with empirical formula CH2OCH_2O and molar mass 180 g/mol?

Answer: C6H12O6C_6H_{12}O_6. 180÷30=6180 \div 30 = 6, so multiply CH2OCH_2O by 6.

Flashcard 17: What is the concentration of a solution with 5 moles of solute in 2 L of solution?

Answer: 2.5 M. 5 moles2 L=2.5\frac{5 \text{ moles}}{2 \text{ L}} = 2.5 M concentration.

Flashcard 18: What is the first step in a stoichiometry problem?

Answer: Write a balanced chemical equation. Ensures atom conservation for stoichiometric calculations.

Flashcard 19: Determine the empirical formula for a compound with 40% carbon, 6.7% hydrogen, and 53.3% oxygen.

Answer: CH2OCH_2O. Convert percentages to moles, then find simplest ratio.

Flashcard 20: What is the excess reactant in 2H2+O22H2O2H_2 + O_2 \rightarrow 2H_2O with 5 moles H2H_2 and 2 moles O2O_2?

Answer: O2O_2. Remains after limiting reactant is completely consumed.

Flashcard 21: Determine the limiting reactant in 2H2+O22H2O2H_2 + O_2 \rightarrow 2H_2O given 5 moles H2H_2 and 2 moles O2O_2.

Answer: H2H_2. Needs 2.5 moles O2O_2, but only 2 available, so H2H_2 limits.

Flashcard 22: How is actual yield different from theoretical yield?

Answer: Actual yield is the measured amount of product obtained. Theoretical assumes perfect conditions; actual is experimental.

Flashcard 23: How is theoretical yield calculated?

Answer: Use stoichiometry from the balanced equation. Assumes complete reaction with limiting reactant consumed.

Flashcard 24: What is the formula to calculate the number of particles from moles?

Answer: particles=moles×6.022×1023\text{particles} = \text{moles} \times 6.022 \times 10^{23}. Multiplies moles by Avogadro's number for particle count.

Flashcard 25: Find the moles of H2OH_2O in 36 grams.

Answer: 2 moles. 36÷18=236 \div 18 = 2 using H2OH_2O molar mass of 18 g/mol.

Flashcard 26: What is the stoichiometric coefficient for H2H_2 in 2H2+O22H2O2H_2 + O_2 \rightarrow 2H_2O?

Answer:

  1. Coefficient shows 2 molecules react in balanced equation.

Flashcard 27: Identify the limiting reactant in 4 moles N2N_2 and 10 moles H2H_2 for N2+3H22NH3N_2 + 3H_2 \rightarrow 2NH_3.

Answer: H2H_2. Need 12 moles H2H_2 but only have 10; H2H_2 limits.

Flashcard 28: How do you calculate mass from moles?

Answer: mass (g)=moles×molar mass (g/mol)\text{mass (g)} = \text{moles} \times \text{molar mass (g/mol)}. Multiplies moles by molar mass to get mass in grams.

Flashcard 29: Find the volume of gas at STP for 1 mole.

Answer: 22.4 L. Standard molar volume at 0°C and 1 atm pressure.

Flashcard 30: What is the percent yield if the theoretical yield is 100g and actual yield is 80g?

Answer: 80%. 80100×100%=80%\frac{80}{100} \times 100\% = 80\% efficiency.

Flashcard 31: Calculate the number of molecules in 2 moles of H2H_2.

Answer: 1.204×10241.204 \times 10^{24} molecules. 2×6.022×10232 \times 6.022 \times 10^{23} molecules per mole.

Flashcard 32: What is the molar mass of H2SO4H_2SO_4?

Answer: 98.1 g/mol. 2H (2) + S (32) + 4O (64) = 98 g/mol total.

Flashcard 33: Convert 3 moles of NaClNaCl to grams.

Answer: 175.5 grams. 3×58.5=175.53 \times 58.5 = 175.5 using NaClNaCl molar mass.

Flashcard 34: Calculate the percent composition of oxygen in H2OH_2O.

Answer: 88.81%. Oxygen mass (16) divided by H2OH_2O mass (18) × 100%.

Flashcard 35: How do you calculate mass from moles?

Answer: mass (g)=moles×molar mass (g/mol)\text{mass (g)} = \text{moles} \times \text{molar mass (g/mol)}. Multiplies moles by molar mass to get mass in grams.

Flashcard 36: What is the stoichiometric coefficient for H2H_2 in 2H2+O22H2O2H_2 + O_2 \rightarrow 2H_2O?

Answer:

  1. Coefficient shows 2 molecules react in balanced equation.

Flashcard 37: Determine the empirical formula for a compound with 40% carbon, 6.7% hydrogen, and 53.3% oxygen.

Answer: CH2OCH_2O. Convert percentages to moles, then find simplest ratio.

Flashcard 38: Determine the moles of NaOHNaOH needed to neutralize 11 mole of HClHCl.

Answer: 11 mole. 1:11:1 mole ratio from balanced acid-base equation.

Flashcard 39: Determine the limiting reactant in 2H2+O22H2O2H_2 + O_2 \rightarrow 2H_2O given 5 moles H2H_2 and 2 moles O2O_2.

Answer: H2H_2. Needs 2.5 moles O2O_2, but only 2 available, so H2H_2 limits.

Flashcard 40: Define molar mass.

Answer: The mass of one mole of a substance in grams. Equals the atomic/molecular weight expressed in g/mol.

Flashcard 41: State Avogadro's number.

Answer: 6.022×10236.022 \times 10^{23}. The number of particles in one mole of any substance.

Flashcard 42: Find the volume of gas at STP for 1 mole.

Answer: 22.4 L. Standard molar volume at 0°C and 1 atm pressure.

Flashcard 43: What is the first step in a stoichiometry problem?

Answer: Write a balanced chemical equation. Ensures atom conservation for stoichiometric calculations.

Flashcard 44: What is the molecular formula for a compound with empirical formula CH2OCH_2O and molar mass 180 g/mol?

Answer: C6H12O6C_6H_{12}O_6. 180÷30=6180 \div 30 = 6, so multiply CH2OCH_2O by 6.

Flashcard 45: Calculate the mass of 0.5 moles of C6H12O6C_6H_{12}O_6.

Answer: 90 grams. 0.5×180=900.5 \times 180 = 90 using glucose molar mass.

Flashcard 46: How is theoretical yield calculated?

Answer: Use stoichiometry from the balanced equation. Assumes complete reaction with limiting reactant consumed.

Flashcard 47: What is the concentration of a solution with 5 moles of solute in 2 L of solution?

Answer: 2.5 M. 5 moles2 L=2.5\frac{5 \text{ moles}}{2 \text{ L}} = 2.5 M concentration.

Flashcard 48: Identify the unit for molar mass.

Answer: grams per mole (g/mol). Standard unit expressing mass per mole of substance.

Flashcard 49: Find the moles of H2OH_2O in 36 grams.

Answer: 2 moles. 36÷18=236 \div 18 = 2 using H2OH_2O molar mass of 18 g/mol.

Flashcard 50: What is the excess reactant in 2H2+O22H2O2H_2 + O_2 \rightarrow 2H_2O with 5 moles H2H_2 and 2 moles O2O_2?

Answer: O2O_2. Remains after limiting reactant is completely consumed.

Flashcard 51: How do you calculate mass from moles?

Answer: mass (g)=moles×molar mass (g/mol)\text{mass (g)} = \text{moles} \times \text{molar mass (g/mol)}. Multiplies moles by molar mass to get mass in grams.

Flashcard 52: Determine the moles of NaOHNaOH needed to neutralize 1 mole of HClHCl.

Answer: 1 mole. 1:1 mole ratio from balanced acid-base equation.

Flashcard 53: Convert 0.25 moles of MgMg to atoms.

Answer: 1.505×10231.505 \times 10^{23} atoms. 0.25×6.022×10230.25 \times 6.022 \times 10^{23} atoms per mole.

Flashcard 54: Calculate the number of molecules in 2 moles of H2H_2.

Answer: 1.204×10241.204 \times 10^{24} molecules. 2×6.022×10232 \times 6.022 \times 10^{23} molecules per mole.

Flashcard 55: What is empirical formula?

Answer: The simplest whole-number ratio of atoms. Shows simplest ratio without molecular complexity.

Flashcard 56: What is the percent yield if the theoretical yield is 100g and actual yield is 80g?

Answer: 80%. 80100×100%=80%\frac{80}{100} \times 100\% = 80\% efficiency.

Flashcard 57: Which unit is used to express concentration in stoichiometry?

Answer: Molarity (M). Standard concentration unit for solution calculations.

Flashcard 58: Find the moles of H2OH_2O in 36 grams.

Answer: 2 moles. 36÷18=236 \div 18 = 2 using H2OH_2O molar mass of 18 g/mol.

Flashcard 59: What is the formula for molarity?

Answer: Molarity (M)=moles of soluteliters of solution\text{Molarity (M)} = \frac{\text{moles of solute}}{\text{liters of solution}}. Relates amount of solute to solution volume.

Flashcard 60: Convert 3 moles of NaClNaCl to grams.

Answer: 175.5 grams. 3×58.5=175.53 \times 58.5 = 175.5 using NaClNaCl molar mass.

Flashcard 61: Calculate the volume at STP for 0.5 moles of gas.

Answer: 11.2 L. 0.5×22.4=11.20.5 \times 22.4 = 11.2 liters at STP.

Flashcard 62: Find the molar mass of CO2CO_2.

Answer: 44 g/mol. Carbon (12) + 2 oxygen atoms (16 each) = 44 g/mol.

Flashcard 63: Convert 3 moles of NaClNaCl to grams.

Answer: 175.5 grams. 3×58.5=175.53 \times 58.5 = 175.5 using NaClNaCl molar mass.

Flashcard 64: Calculate the mass of 0.5 moles of C6H12O6C_6H_{12}O_6.

Answer: 90 grams. 0.5×180=900.5 \times 180 = 90 using glucose molar mass.

Flashcard 65: What is the formula to calculate the number of moles from mass?

Answer: moles=mass (g)molar mass (g/mol)\text{moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}}. Divides mass by molar mass to find amount in moles.

Flashcard 66: What is empirical formula?

Answer: The simplest whole-number ratio of atoms. Shows simplest ratio without molecular complexity.

Flashcard 67: Determine the limiting reactant in 2H2+O22H2O2H_2 + O_2 \rightarrow 2H_2O given 5 moles H2H_2 and 2 moles O2O_2.

Answer: H2H_2. Needs 2.5 moles O2O_2, but only 2 available, so H2H_2 limits.

Flashcard 68: What is the concentration of a solution with 5 moles of solute in 2 L of solution?

Answer: 2.5 M. 5 moles2 L=2.5\frac{5 \text{ moles}}{2 \text{ L}} = 2.5 M concentration.

Flashcard 69: Calculate the mass of 0.5 moles of C6H12O6C_6H_{12}O_6.

Answer: 90 grams. 0.5×180=900.5 \times 180 = 90 using glucose molar mass.

Flashcard 70: How is actual yield different from theoretical yield?

Answer: Actual yield is the measured amount of product obtained. Theoretical assumes perfect conditions; actual is experimental.

Flashcard 71: How many moles of H2H_2 are needed to produce 2 moles of NH3NH_3 in N2+3H22NH3N_2 + 3H_2 \rightarrow 2NH_3?

Answer: 3 moles. 3:2 mole ratio from balanced equation coefficients.

Flashcard 72: How is theoretical yield calculated?

Answer: Use stoichiometry from the balanced equation. Assumes complete reaction with limiting reactant consumed.

Flashcard 73: Identify the unit for molar mass.

Answer: grams per mole (g/mol). Standard unit expressing mass per mole of substance.

Flashcard 74: Find the volume of gas at STP for 1 mole.

Answer: 22.4 L. Standard molar volume at 0°C and 1 atm pressure.

Flashcard 75: Determine the empirical formula for a compound with 40% carbon, 6.7% hydrogen, and 53.3% oxygen.

Answer: CH2OCH_2O. Convert percentages to moles, then find simplest ratio.

Flashcard 76: Find the volume of 3 moles of gas at STP.

Answer: 67.2 L. 3×22.4=67.23 \times 22.4 = 67.2 liters at standard conditions.

Flashcard 77: Identify the unit for molar mass.

Answer: grams per mole (g/mol). Standard unit expressing mass per mole of substance.

Flashcard 78: What is the formula for molarity?

Answer: Molarity (M)=moles of soluteliters of solution\text{Molarity (M)} = \frac{\text{moles of solute}}{\text{liters of solution}}. Relates amount of solute to solution volume.

Flashcard 79: How is theoretical yield calculated?

Answer: Use stoichiometry from the balanced equation. Assumes complete reaction with limiting reactant consumed.

Flashcard 80: What is the excess reactant in 2H2+O22H2O2H_2 + O_2 \rightarrow 2H_2O with 5 moles H2H_2 and 2 moles O2O_2?

Answer: O2O_2. Remains after limiting reactant is completely consumed.

Flashcard 81: What is the excess reactant in 2H2+O22H2O2H_2 + O_2 \rightarrow 2H_2O with 5 moles H2H_2 and 2 moles O2O_2?

Answer: O2O_2. Remains after limiting reactant is completely consumed.

Flashcard 82: How do you calculate mass from moles?

Answer: mass (g)=moles×molar mass (g/mol)\text{mass (g)} = \text{moles} \times \text{molar mass (g/mol)}. Multiplies moles by molar mass to get mass in grams.

Flashcard 83: What is the molecular formula for a compound with empirical formula CH2OCH_2O and molar mass 180 g/mol?

Answer: C6H12O6C_6H_{12}O_6. 180÷30=6180 \div 30 = 6, so multiply CH2OCH_2O by 6.

Flashcard 84: Determine the moles of NaOHNaOH needed to neutralize 1 mole of HClHCl.

Answer: 1 mole. 1:11:1 mole ratio from balanced acid-base equation.

Flashcard 85: Determine the empirical formula for a compound with 50% sulfur and 50% oxygen by mass.

Answer: SO2SO_2. Equal mass percentages give 1:2 atomic ratio S:O.

Flashcard 86: Identify the limiting reactant in 4 moles N2N_2 and 10 moles H2H_2 for N2+3H22NH3N_2 + 3H_2 \rightarrow 2NH_3.

Answer: H2H_2. Need 12 moles H2H_2 but only have 10; H2H_2 limits.

Flashcard 87: What is the formula for percent yield?

Answer: Percent yield=actual yieldtheoretical yield×100\text{Percent yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100. Compares actual to theoretical yield as a percentage.

Flashcard 88: Determine the limiting reactant in 2H2+O22H2O2H_2 + O_2 \rightarrow 2H_2O given 5 moles H2H_2 and 2 moles O2O_2.

Answer: H2H_2. Needs 2.5 moles O2O_2, but only 2 available, so H2H_2 limits.

Flashcard 89: What is the formula to calculate the number of moles from mass?

Answer: moles=mass (g)molar mass (g/mol)\text{moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}}. Divides mass by molar mass to find amount in moles.

Flashcard 90: What is the stoichiometric coefficient for H2H_2 in 2H2+O22H2O2H_2 + O_2 \rightarrow 2H_2O?

Answer:

  1. Coefficient shows 2 molecules react in balanced equation.

Flashcard 91: What is the formula for molarity?

Answer: Molarity (M)=moles of soluteliters of solution\text{Molarity (M)} = \frac{\text{moles of solute}}{\text{liters of solution}}. Relates amount of solute to solution volume.

Flashcard 92: What is the formula to calculate the number of particles from moles?

Answer: particles=moles×6.022×1023\text{particles} = \text{moles} \times 6.022 \times 10^{23}. Multiplies moles by Avogadro's number for particle count.

Flashcard 93: How many grams of O2O_2 are needed to react with 4 moles of CC in C+O2CO2C + O_2 \rightarrow CO_2?

Answer: 128 grams. 4×32=1284 \times 32 = 128 grams using 1:1 mole ratio.

Flashcard 94: What is the formula to calculate the number of moles from mass?

Answer: moles=mass (g)molar mass (g/mol)\text{moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}}. Divides mass by molar mass to find amount in moles.

Flashcard 95: How many moles of H2H_2 are needed to produce 2 moles of NH3NH_3 in N2+3H22NH3N_2 + 3H_2 \rightarrow 2NH_3?

Answer: 3 moles. 3:2 mole ratio from balanced equation coefficients.

Flashcard 96: What is empirical formula?

Answer: The simplest whole-number ratio of atoms. Shows simplest ratio without molecular complexity.

Flashcard 97: What is the percent yield if the theoretical yield is 100g and actual yield is 80g?

Answer: 80%. 80100×100%=80%\frac{80}{100} \times 100\% = 80\% efficiency.

Flashcard 98: Calculate the mass of 0.5 moles of C6H12O6C_6H_{12}O_6.

Answer: 90 grams. 0.5×180=900.5 \times 180 = 90 using glucose molar mass.

Flashcard 99: State Avogadro's number.

Answer: 6.022×10236.022 \times 10^{23}. The number of particles in one mole of any substance.

Flashcard 100: What is the formula for molarity?

Answer: Molarity (M)=moles of soluteliters of solution\text{Molarity (M)} = \frac{\text{moles of solute}}{\text{liters of solution}}. Relates amount of solute to solution volume.