Question

Silver nitrate reacts with calcium chloride as

Answer 1

Answer: 11.8 grams of $CaCl_2$ is required.

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in L)}}$     .....(1)

Molarity of $AgNO_3$ solution = 0.506 M

Volume of solution = 420 mL

Putting values in equation 1, we get:

$0.506M=\frac{\text{Moles of} AgNO_3\times 1000}{420ml}\\\\\text{Moles of }AgNO_3=\frac{0.506mol/L\times 420}{1000}=0.212mol$

The balanced chemical reaction is:

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

2 mole of $AgNO_3$ requires = 1 mole of $CaCl_2$  

0.212 moles of $AgNO_3$ require = $\frac{1}{2}\times 0.212=0.106$ moles of $CaCl_2$

Mass of  $CaCl_2$ =$moles\times {\text {Molar Mass}}=0.106mol\times 111g/mol=11.8g$