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

Question

asked Dec 8, 2021 80.1k views

1 Answer

Avand Amiri · Answer 1 · 2021-12-13T04:32:26+0000

Answer:

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 ,

$m = mol * molar \ mass$

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