Question

How many grams of Ag2CO3 will precipitate when excess Na2CO3 solution is added to 40.0 mL of 0.679 M AgNO3 solution

Answer 1

The answer is "
`37.45364 \ g`"

Step-by-step explanation:

Equation:

`2AgNO_3 (aq) + Na_2CO_3 (aq) \longrightarrow Ag_2CO_3 (s) + 2NaNO_3 (aq)`

Calculating the mol of
`AgNO_3`:

`\to V = 40.0 mL\\\\`

`= 40.0 * 10^(-2) \ L \\\\`

`\to n = Molarity * Volume \\\\`

`= 0.679 * 40.0 * 10^(-2)\\\\= 27.16 * 10^(-2) \ mol`

mol of
`Ag_2CO_3= (1)/(2) * n\\\\`

`= (1)/(2) * 27.16 * 10^(-2)\\\\= 13.58 * 10^(-2)\\\\`

Calculating the molar mass of
`Ag_2CO_3`,:

`= 2* MM(Ag) + 1 * MM(C) + 3 * MM(O)\\\\= 2 * 107.9 + 1 * 12.01 + 3 * 16.0\\\\= 275.81 \ (g)/(mol)`

Calculating the mass of
`Ag_2CO_3`,

`m = mol * molar \ mass`

`= 13.38 * 10^(-2)\ mol * 2.758 * 10^2 \ (g)/(mol)\\\\= 37.45364 \ g`