Question

When solutions of silver nitrate and sodium chloride are mixed, a precipitation reaction occurs. What mass of precipitate can be produced from 1.14 L of a 0.269 M solution of silver nitrate reacting with excess sodium chloride?

Answer 1

Explanation:

The given precipitation reaction will be as follows.

        $AgNO_{3}(aq) + NaCl(aq) \rightarrow AgCl(s) + NaNO_{3}(aq)$

Here, AgCl is the precipitate which is formed.

It is known that molarity is the number of moles present in a liter of solution.

Mathematically,       Molarity = $\frac{\text{no. of moles}}{\text{volume in liter}}$

It is given that volume is 1.14 L and molarity is 0.269 M. Therefore, calculate number of moles as follows.

                    Molarity = $\frac{\text{no. of moles}}{\text{volume in liter}}$

                     0.269 M = $\frac{\text{no. of moles}}{1.14 L}$

                    no. of moles = 0.306 mol

As molar mass of AgCl is 143.32 g/mol. Also, relation between number of moles and mass is as follows.

               No. of moles = $\frac{mass}{\text{molar mass}}$

                   0.307 mol = $\frac{mass}{143.32 g/mol}$

                           mass = 43.99 g

Thus, we can conclude that mass of precipitate produced is 43.99 g.