Question

What mass of calcium chloride in grams would be enough to produce 14.35 gm of AgCl

Answer 1

11.11gm mass of calcium chloride would be enough to produce 14.35gm of agcl.

If any doubt ask it in comment box

Answer 2

Answer: 3.7827 grams of calcium chloride.

Explanation:

$2AgNO_3+CaCl_2\rightarrow 2AgCl+Ca(NO_3)_2$

moles of AgCl =$\frac{\text{mass of AgCl}}{\text{molar mass of AgCl}}=\frac{14.35 g}{143.32 g/mol}=0.1001 mol$

According to reaction, 2 moles of AgCl are produced from 1 mole of $CaCl_2$ then 0.1001 mole of AgCl will produce:$\frac{1}{2}\times 0.1001$ moles of  $CaCl_2$ that is 0.05005 moles.

Mass of $CaCl_2$ =

moles of $CaCl_2$ × molar mass of   $CaCl_2$

= 0.05005 moles × 75.58g/mol = 3.7827 grams

3.7827 grams of calcium chloride would be enough to produce 14.35 g of AgCl.