#### Answer

7.97 g of AgCl

#### Work Step by Step

- 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:
$$ 0.0556 \space mole \times \frac{ 143.4 \space g}{1 \space mole} = 7.97 \space g$$