In order to determine the limiting reactant, we can use a calculation to determine how much of one reactant is needed in order to use up the other. For example, we can see how much hydrogen gas is necessary in order to use up all \(\displaystyle \small 50g\) of nitrogen gas that we have:

\(\displaystyle 50gN_{2} * \frac{1molN_{2}}{28gN_{2}} * \frac{3molesH_{2}}{1molN_{2}}* \frac{2.02gH_{2}}{1molH_{2}} = 11 gramsH_{2}\)

In other words, we need \(\displaystyle \small 11g\) of hydrogen gas in order to use up \(\displaystyle \small 50g\) of nitrogen gas. Since we have \(\displaystyle \small 15g\) of hydrogen gas, we have more than enough to react all of the nitrogen gas, and the nitrogen gas will be used up before the hydrogen gas. As a result, nitrogen gas is the limiting reactant.

Limiting reactants are determined when there is an excess of a particular reactant, in relation to the other reactants available. When reactants are compared, one must always compare the molar ratio in order to determine which reactant is limited. In this reaction, if we are given the available amount of each reactant, we will need to convert the given amount of one reactant (A) to the necessary amount of the other (B) required to fully react. If we find that reactant A is available in excess, then reactant B will be the limiting reagent. The only way to compare these two terms, however, is by using the molar ratio in the reaction.

Let's begin by converting our reactants from grams to moles. Note that both gases are diatomic, meaning that there are two atoms per molecule.

Molecular weights:

\(\displaystyle 2(O)=2(16)=32\frac{g}{mol}\)

\(\displaystyle 2(H)=2(1)=2\frac{g}{mol}\)

Conversion to moles:

\(\displaystyle 64g\ O_2*\frac{1mol}{32g}=2mol\ O_2\)

\(\displaystyle 6g\ H_2*\frac{1mol}{2g}=3mol\ H_2\)

We have 2 moles of oxygen gas and 3 moles of hydrogen gas. Next, look at the given reaction:

\(\displaystyle 2H_2+O_2\rightarrow2H_2O\)

Two moles of hydrogen gas are consumed to react each mole of oxygen gas. We can use this ratio to find the number of moles of oxygen gas needed to react all of the hydrogen given in this question.

\(\displaystyle 3mol\ H_2*\frac{1mol\ O_2}{2mol\ H_2}=1.5mol\ O_2\)

Only 1.5 moles of oxygen gas are needed to react all of the hydrogen gas. Since we have 2 moles of oxygen gas, some of it will remain unreacted when all of the hydrogen gas has been used. This means the hydrogen gas is the limiting reagent.

Find the amount of unreacted oxygen gas:

\(\displaystyle 2mol\ O_2-1.5mol\ O_2=0.5mol\ O_2\)

Convert to grams:

\(\displaystyle 0.5mol\ O_2*\frac{32g}{1mol}=16g\ O_2\)

First, we must determine how many moles of each reactant begin the reaction by multiplying the molarity by the volume. Don't forget to convert volume to liters!

\(\displaystyle mol=M*V\)

\(\displaystyle (0.250M)(0.050L)=0.0125mol\ K_2CO_3\)

\(\displaystyle (0.175M)(0.100L)=0.0175mol\ Ca(NO_3)_2\)

Next, use the reaction coefficients (i.e. the stoichiometry) to determine how many moles of calcium carbonate could be formed from each of the reactants. In this case, there is a 1:1 molar ratio between both reactants and calcium carbonate.

\(\displaystyle 0.0125mol\ K_2CO_3*\frac{1mol\ CaCO_3}{1mol\ K_2CO_3}=0.0125mol\ CaCO_3\)

\(\displaystyle 0.0175mol\ Ca(NO_3)_2*\frac{1mol\ CaCO_3}{1mol\ Ca(NO_3)_2}=0.0175mol\ CaCO_3\)

Thus, 0.0125 moles of potassium carbonate could form 0.0125 moles of calcium carbonate, while 0.0175 moles of calcium nitrate could form 0.0175 moles of calcium carbonate. The maximum amount of product is going to be determined by the limiting reactant, i.e. the reactant that provides the least amount of product. In this case, the limiting reactant is potassium carbonate, and the maximum yield of calcium carbonate is 0.0125mol.

For the final step convert this value to grams:

\(\displaystyle 0.0125mol\ CaCO_3*\frac{100.01g\ CaCO_3}{1mol\ CaCO_3}=1.25g\ CaCO_3\)

First, we must determine how many moles of each reactant begin the reaction by multiplying the molarity by the volume. Don't forget to convert volume to liters!

\(\displaystyle mol=M*V\)

\(\displaystyle (0.250M)(0.050L)=0.0125mol\ K_2CO_3\)

\(\displaystyle (0.175M)(0.100L)=0.0175mol\ Ca(NO_3)_2\)

Next, use the reaction coefficients (i.e. the stoichiometry) to determine how many moles of calcium carbonate could be formed from each of the reactants. In this case, there is a 1:1 molar ratio between both reactants and calcium carbonate.

\(\displaystyle 0.0125mol\ K_2CO_3*\frac{1mol\ CaCO_3}{1mol\ K_2CO_3}=0.0125mol\ CaCO_3\)

\(\displaystyle 0.0175mol\ Ca(NO_3)_2*\frac{1mol\ CaCO_3}{1mol\ Ca(NO_3)_2}=0.0175mol\ CaCO_3\)

Thus, 0.0125 moles of potassium carbonate could form 0.0125 moles of calcium carbonate, while 0.0175 moles of calcium nitrate could form 0.0175 moles of calcium carbonate. The maximum amount of product is going to be determined by the limiting reactant, i.e. the reactant that provides the least amount of product. In this case, the limiting reactant is potassium carbonate, and the maximum yield of calcium carbonate is 0.0125mol.

For the final step convert this value to grams:

\(\displaystyle 0.0125mol\ CaCO_3*\frac{100.01g\ CaCO_3}{1mol\ CaCO_3}=1.25g\ CaCO_3\)

The remaining mass with be equal to the mass of the sodium chloride in the solution. Once the solvent (water) evaporates, the solute will remain.

Atomic mass of sodium is ~23. Atomic mass of chlorine is ~35.5. Molecular mass of NaCl is ~58.5.

\(\displaystyle 0.100L*\frac{0.4mol}{1L}=0.040mol\ NaCl\)

\(\displaystyle 0.040mol\ NaCl*\frac{58.5g\ NaCl}{1mol\ NaCl}=2.34g\ NaCl\)

This question is asking us to solve for molality, which has the following equation:

\(\displaystyle m=\frac{\text{moles solute}}{\text{kg solvent}}\)

Note how the equation does not ask for liters of solution like molarity, but instead requires the kilograms of solvent used in the solution. By dividing the mass of the sodium chloride by its molar mass, we can solve for the moles of sodium chloride in solution.

\(\displaystyle NaCl\)

\(\displaystyle MM=(23)+(35.45)=58.45\frac{g}{mol}\)

\(\displaystyle moles=\frac{mass}{molar mass}=\frac{100g}{58.45\frac{g}{mol}}=1.71mol\ NaCl\)

This will allow us to solve for the mass of water used in the solution.

\(\displaystyle m=\frac{1.71mol}{x}=3.0m\)

\(\displaystyle x=\frac{1.71mol}{3.0m}=0.57kg=570g\)

This gives us the mass of water required to make the solution.

First, we must determine how many moles of each reactant begin the reaction by multiplying the molarity by the volume. Don't forget to convert volume to liters!

\(\displaystyle mol=M*V\)

\(\displaystyle (0.250M)(0.050L)=0.0125mol\ K_2CO_3\)

\(\displaystyle (0.175M)(0.100L)=0.0175mol\ Ca(NO_3)_2\)

Next, use the reaction coefficients (i.e. the stoichiometry) to determine how many moles of calcium carbonate could be formed from each of the reactants. In this case, there is a 1:1 molar ratio between both reactants and calcium carbonate.

\(\displaystyle 0.0125mol\ K_2CO_3*\frac{1mol\ CaCO_3}{1mol\ K_2CO_3}=0.0125mol\ CaCO_3\)

\(\displaystyle 0.0175mol\ Ca(NO_3)_2*\frac{1mol\ CaCO_3}{1mol\ Ca(NO_3)_2}=0.0175mol\ CaCO_3\)

Thus, 0.0125 moles of potassium carbonate could form 0.0125 moles of calcium carbonate, while 0.0175 moles of calcium nitrate could form 0.0175 moles of calcium carbonate. The maximum amount of product is going to be determined by the limiting reactant, i.e. the reactant that provides the least amount of product. In this case, the limiting reactant is potassium carbonate, and the maximum yield of calcium carbonate is 0.0125mol.

For the final step convert this value to grams:

\(\displaystyle 0.0125mol\ CaCO_3*\frac{100.01g\ CaCO_3}{1mol\ CaCO_3}=1.25g\ CaCO_3\)