Question

Consider the following unbalanced equation: AgNO3 (aq) CaCl2 (aq) → AgCl (s) Ca(NO3)2 (aq) How many grams of silver chloride can be prepared by the reaction of 100.0 mL of 0.20 M silver nitrate with 100.0 mL of 0.15 M calcium chloride?

Answer 1

Answer: 3 grams

Explanation:-

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

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

Molarity of $AgNO_3$ solution = 0.20 M

Volume of $AgNO_3$ solution = 100.0 mL = 0.1 L

Putting values in equation 1, we get:

$\text{Moles of} AgNO_3={0.20}\times{0.1}=0.02moles$

Molarity of $CaCl_2$ solution = 0.15 M

Volume of $CaCl_2$ solution = 100.0 mL = 0.1 L

Putting values in equation 1, we get:

$\text{Moles of} CaCl_2={0.15}\times{0.1}=0.015moles$

The balanced chemical equation for the reaction is:

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

According to stoichiometry :

2 moles of $AgNO_3$ require 1 mole of $CaCl_2$

Thus 0.02 moles of $AgNO_3$ will require=$\frac{1}{2}\times 0.02=0.01moles$  of $CaCl_2$

Thus $AgNO_3$ is the limiting reagent as it limits the formation of product and $CaCl_2$ is the excess reagent.

As 2 mole of $AgNO_3$ give = 2 mole of $AgCl$

Thus 0.02 moles of $AgNO_3$ give =$\frac{1}{1}\times 0.02=0.02moles$  of $AgCl$

Mass of $AgCl=moles\times {\text {Molar mass}}=0.02moles\times 143.32g/mol=3g$

Thus 3 g of $AgCl$ will be produced from the given masses of both reactants.