Answer:
The answer is "[tex]37.45364 \ g[/tex]"
Explanation:
Equation:
[tex]2AgNO_3 (aq) + Na_2CO_3 (aq) \longrightarrow Ag_2CO_3 (s) + 2NaNO_3 (aq)[/tex]
Calculating the mol of [tex]AgNO_3[/tex]:
[tex]\to V = 40.0 mL\\\\[/tex]
[tex]= 40.0 \times 10^{-2} \ L \\\\[/tex]
[tex]\to n = Molarity \times Volume \\\\[/tex]
[tex]= 0.679 \times 40.0 \times 10^{-2}\\\\= 27.16 \times 10^{-2} \ mol[/tex]
mol of [tex]Ag_2CO_3= \frac{1}{2} \times n\\\\[/tex]
[tex]= \frac{1}{2} \times 27.16 \times 10^{-2}\\\\= 13.58 \times 10^{-2}\\\\[/tex]
Calculating the molar mass of [tex]Ag_2CO_3[/tex],:
[tex]= 2\times MM(Ag) + 1 \times MM(C) + 3 \times MM(O)\\\\= 2 \times 107.9 + 1 \times 12.01 + 3 \times 16.0\\\\= 275.81 \ \frac{g}{mol}[/tex]
Calculating the mass of [tex]Ag_2CO_3[/tex],
[tex]m = mol \times molar \ mass[/tex]
[tex]= 13.38 \times 10^{-2}\ mol \times 2.758 \times 10^2 \ \frac{g}{mol}\\\\= 37.45364 \ g[/tex]
