## Chemistry 10th Edition

- Calculate or find the molar mass for $AgNO_3$: $AgNO_3$ : ( 107.9 $\times$ 1 )+ ( 14.01 $\times$ 1 )+ ( 16.00 $\times$ 3 )= 169.9 g/mol - Using the molar mass as a conversion factor, find the amount in moles: $$9.45 \space g \times \frac{1 \space mole}{ 169.9 \space g} = 0.0556 \space mole$$ - Calculate or find the molar mass for $CaCl_2$: $CaCl_2$ : ( 35.45 $\times$ 2 )+ ( 40.08 $\times$ 1 )= 110.98 g/mol - Using the molar mass as a conversion factor, find the amount in moles: $$6.30 \space g \times \frac{1 \space mole}{ 110.98 \space g} = 0.0568 \space mole$$ Find the amount of product if each reactant is completely consumed. $$0.0556 \space mole \space AgNO_3 \times \frac{ 2 \space moles \ AgCl }{ 2 \space moles \space AgNO_3 } = 0.0556 \space mole \space AgCl$$ $$0.0568 \space mole \space CaCl_2 \times \frac{ 2 \space moles \ AgCl }{ 1 \space mole \space CaCl_2 } = 0.114 \space mole \space AgCl$$ Since the reaction of $AgNO_3$ produces less $AgCl$ for these quantities, it is the limiting reactant. - Calculate or find the molar mass for $AgCl$: $AgCl$ : ( 35.45 $\times$ 1 )+ ( 107.9 $\times$ 1 )= 143.4 g/mol - Using the molar mass as a conversion factor, find the mass in g: $$0.0556 \space mole \times \frac{ 143.4 \space g}{1 \space mole} = 7.97 \space g$$