The chemical reaction that would occur is provided below:

$\displaystyle AgNO_{3(aq)}+NaCl_{(aq)}\rightarrow NaNO_{3(aq)} + AgCl_{(s)}$

This type of reaction is called a double displacement reaction in which there is an exchange of ions between two compounds. This type of reaction usually results in the formation of a precipitate or gas. In terms of the positive ions, the $\displaystyle Ag^{+}$ switch places $\displaystyle Na^{+}$ to form two new products.

The type of reaction given is termed a combustion reaction. A combustion reaction is one that involves a substance reacting with molecular oxygen. For combustion reactions involving a hydrocarbon as given in the example, the major products are carbon dioxide and water. The word combustion is a synonym for burn.

A displacement reaction is one in which an atom, ion or molecule moves from one compound to another to replace an atom, ion or molecule in another compound. In the reaction given, $\displaystyle SO_{4}^{2-}$ ion moves from $\displaystyle CuSO_{4}$ to $\displaystyle Zn(s)$ to form $\displaystyle ZnSO_{4}$.

In chemistry, percentage yield is a term used to quantify the efficiency of a chemical reaction. In order to calculate this quantity, the value for the actual yield and theoretical yield is needed. The actual yield is the amount of product obtained during a chemical reaction. The theoretical yield is the maximum amount of product possible to be obtain by a chemical reaction which is based on the amount of reactants used. Below is the formula needed to calculate the percentage yield.

Convert the number of grams of $\displaystyle Ba(NO_{3})_{2}$ to moles:

$\displaystyle 12.6 g\ Ba(NO_{3})_{2}\times \frac{mole}{261.37g}=0.0482\ moles\ Ba(NO_{3})_{2}$

Based on the chemical equation 1 mole of $\displaystyle BaSO_{4}$ is produced for every 1 mole of $\displaystyle Ba(NO_{3})_{2}$ reacted.

$\displaystyle moles\ of Ba(NO_{3})_{2}=moles\ of\ BaSO_{4}$

Convert the moles of $\displaystyle BaSO_{4}$ to grams:

$\displaystyle 0.0482\ moles\ BaSO_{4} \times\ \frac{233.43g}{moles}=11.3g\ BaSO_{4}$

$\displaystyle Percentage\ Yield =\frac{actual\ yield}{theoretical\ yield}\times100$

$\displaystyle Percentage\ Yield =\frac{10.4g}{11.3g}\times100=92.0 \%$

In chemistry, percentage yield is a term used to quantify the efficiency of a chemical reaction. In order to calculate this quantity, the value for the actual yield and theoretical yield is needed. The actual yield is the amount of product obtained during a chemical reaction. The theoretical yield is the maximum amount of product possible to be obtain by a chemical reaction which is based on the amount of reactants used. Below is the formula needed to calculate the percentage yield.

Convert the number of grams of $\displaystyle NaOH$ to moles:

$\displaystyle 21.0 g\ NaOH\times \frac{mole}{40.0g}=0.525\ moles\ NaOH$

Based on the chemical equation 1 mole of $\displaystyle NaCl$ is produced for every 1 mole of $\displaystyle NaOH$ reacted.

$\displaystyle moles\ of\ NaOH=moles\ of\ NaCl$

Convert the moles of $\displaystyle NaCl$ to grams:

$\displaystyle 0.525\ moles\ \times\ \frac{58.44g}{moles}=30.7g\ NaCl$

$\displaystyle Percentage\ Yield =\frac{actual\ yield}{theoretical\ yield}\times100$

$\displaystyle Percent\ Yield =\frac{10.2g}{30.7g}\times100=33.2 \%$

Convert the number of grams of $\displaystyle Al(OH)_{3}$ to moles:

$\displaystyle 5.6 g\ Al(OH)_{3}\times \frac{mole}{78.00g}=0.0718\ moles\ Al(OH)_{3}$

Based on the chemical equation 1 mole of $\displaystyle Al_{2}(SO_{4})_{3}$ is produced for every 2 mole of $\displaystyle Al(OH)_{3}$ reacted.

$\displaystyle 2\ moles\ of Al(OH)_{3}=1\ mole\ of\ Al_{2}(SO_{4})_{3}$

Therefore: $\displaystyle 0.0718\ moles\ Al(OH)_{3}\times \frac{1\ moles\ Al_{2}(SO_{4})_{3}}{2\ moles\ Al(OH)_{3}}=0.0359\ moles\ Al_{2}(SO_{4})_{3}$

Convert the moles of $\displaystyle Al_{2}(SO_{4})_{3}$ to grams:

$\displaystyle 0.0359\ moles\ Al_{2}(SO_{4})_{3} \times\ \frac{342.15g}{moles}=12.3g\ Al_{2}(SO_{4})_{3}$

A displacement reaction is one in which an atom, ion or molecule moves from one compound to another to replace an atom, ion or molecule in another compound. In the reaction given, $\displaystyle NO_{3}^{-}$ ion moves from $\displaystyle AgNO_{3}$ to $\displaystyle Na_{2}CO_{3}$. Also, $\displaystyle CO_{3}^{2-}$ion moves from $\displaystyle Na_{2}CO_{3}$ to $\displaystyle AgNO_{3}$.

In chemistry, percentage yield is a term used to quantify the efficiency of a chemical reaction. In order to calculate this quantity, the value for the actual yield and theoretical yield is needed. The actual yield is the amount of product obtained during a chemical reaction. The theoretical yield is the maximum amount of product possible to be obtain by a chemical reaction which is based on the amount of reactants used. Below is the formula needed to calculate the percentage yield.

Convert the number of grams of $\displaystyle CaCO_{3}$ to moles:

$\displaystyle 40.0 g\ CaCO_{3}\times \frac{mole}{100.09g}=0.3996\ moles\ CaCO_{3}$

Based on the chemical equation 1 mole of $\displaystyle CaO$ is produced for every 1 mole of $\displaystyle CaCO_{3}$ reacted.

$\displaystyle moles\ of\ CaCO_{3}=moles\ of\ CaO$

Convert the moles of $\displaystyle CaO$ to grams:

$\displaystyle 0.3996\ moles\ \times\ \frac{56.08g}{moles}=22.0g\ CaO$

$\displaystyle Percentage\ Yield =\frac{actual\ yield}{theoretical\ yield}\times100$

$\displaystyle Percentage\ Yield =\frac{15.0g}{22.0g}\times100=68.2 \%$

A displacement reaction is one in which an atom, ion or molecule moves from one compound to another to replace an atom, ion or molecule in another compound. In the reaction given, 2 $\displaystyle Cl^{-}$ ions move from 2 moles of $\displaystyle HCl$ to $\displaystyle Cu(s)$ to form $\displaystyle CuCl_{2}$.

When given initial concentrations, we can determine the reaction quotient (Q) of the reaction. The reaction quotient is given by the same equation as the equilibrium constant (concentration of products divided by concentration of reactants), but its value will fluctuate as the system reacts, whereas the equilibrium constant is based on equilibrium concentrations.

By comparing the reaction quotient to the equilibrium constant, we can determine in which direction the reaction will proceed initially. 

$\displaystyle Q\ \text{or}\ K_{eq}=\frac{[CH_3OH]}{[H_2]^2[CO]}$

The reaction quotient with the beginning concentrations is written below.

$\displaystyle \frac{[2.5]}{[0.6]^{2}[1.3]} = 5.3$

This shows that the ratio of products to reactants is less than the equilibrium constant. Since Q is less than K_eq in the beginning, we conclude that the reaction will proceed forward until Q is equal to K_eq. In this manner, the denominator (reactants) will decrease and the numerator (products) will increase, causing Q to become closer to K_eq.