Answer:
The mass of silver carbonate precipitated is 5.18 grams.
Explanation:
Molarity of the silver nitrate solution = 0.671 M
Volume of the silver nitrate solution = 56.0 mL
[tex]1 mL = 0.001 L\\56.0 mL = 56.0\times 0.001 L=0.0560 L[/tex]
Moles of silver nitrate = n
[tex]Molarity=\frac{\text{Moles of compound}}{\text{Volume of solution in Liters}}\\\\0.671 M=\frac{n}{0.0560 L}\\n=0.671 M\times 0.0560 L=0.0376 mol[/tex]
Moles of silver nitrate used = 0.0376 mol
[tex]K_2CO_3+2AgNO_3\rightarrow Ag_2CO_3+2KNO_3[/tex]
According to the reaction, 2 moles of silver nitrate gives 1 mole of silver carbonate, then 0.0376 moles of silver nitrate:
[tex]=\frac{1}{2}\times 0.0376 mol=0.0188 \text{mol of }Ag_2CO_3[/tex]
Moles of the silver carbonate formed = 0.0188 mol
Molar mass of silver carbonate = 275.7453 g/mol
Mass of silver carbonate :
[tex]=275.7453 g/mol\times 0.0188 mol=5.1840 g\approx 5.18 g[/tex]
The mass of silver carbonate precipitated is 5.18 grams.
