## Chemistry 10th Edition

(a) $AgNO_3$ is the limiting reactant. (b) 14.9 g of $BaCl_2$ will be left over. (c) 52.7 g of $AgCl$ will be produced.
(a) 1. Calculate the number of moles of $BaCl_2$: 137.3* 1 + 35.45* 2 = 208.2g/mol $53.1g \times \frac{1 mol}{ 208.2g} = 0.255mol (BaCl_2)$ According to the coefficients of the balanced equation: The ratio of $BaCl_2$ to $AgNO_3$ is 1 to 2: $0.255 mol (BaCl_2) \times \frac{ 2 mol(AgNO_3)}{ 1 mol (BaCl_2)} = 0.5101mol (AgNO_3)$ 2. Calculate the number of moles of $AgNO_3$: 107.9* 1 + 14.01* 1 + 16* 3 = 169.9g/mol $62.4g \times \frac{1 mol}{ 169.9g} = 0.3673mol (AgNO_3)$ Therefore, $AgNO_3$ is the limiting reactant, because 53.1g of $BaCl_2$ can react with more than $0.3673mol(AgNO_3)$, so $BaCl_2$ is the excess reactant. ------------- (b) Use the number of moles of the limiting reactant $AgNO_3$: The ratio of $AgNO_3$ to $BaCl_2$ is 2 to 1: $0.3673 mol (AgNO_3) \times \frac{ 1 mol(BaCl_2)}{ 2 mol (AgNO_3)} = 0.1837mol (BaCl_2)$ 2. Calculate the mass of $BaCl_2$ that will react with: 137.3* 1 + 35.45* 2 = 208.2g/mol $0.1837 mol \times \frac{ 208.2 g}{ 1 mol} = 38.2g (BaCl_2)$ - We had 53.1 g of $BaCl_2$, and 38.2 g of it reacted: $53.1 g - 38.2g = 14.9g(Excess-BaCl_2)$ ------------------------------------------ (c) The ratio of $AgNO_3$ to $AgCl$ is 2 to 2: $0.3673 mol (AgNO_3) \times \frac{ 2 mol(AgCl)}{ 2 mol (AgNO_3)} = 0.3673mol (AgCl)$ 2. Calculate the mass of $AgCl$: 107.9* 1 + 35.45* 1 = 143.4g/mol $0.3673 mol \times \frac{ 143.4 g}{ 1 mol} = 52.7g (AgCl)$