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

Step-by-step 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 =
`(mass)/(143.32 g/mol)`

mass = 43.99 g

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