Question

How many grams of Ag2CO3 will precipitate when excess K2CO3 solution is added to 56.0 mL of 0.671 M AgNO3 solution?

Answer 1

The mass of silver carbonate precipitated is 5.18 grams.

Step-by-step explanation:

Molarity of the silver nitrate solution = 0.671 M

Volume of the silver nitrate solution = 56.0 mL

`1 mL = 0.001 L\\56.0 mL = 56.0* 0.001 L=0.0560 L`

Moles of silver nitrate = n

`Molarity=\frac{\text{Moles of compound}}{\text{Volume of solution in Liters}}\\\\0.671 M=(n)/(0.0560 L)\\n=0.671 M* 0.0560 L=0.0376 mol`

Moles of silver nitrate used = 0.0376 mol

`K_2CO_3+2AgNO_3\rightarrow Ag_2CO_3+2KNO_3`

According to the reaction, 2 moles of silver nitrate gives 1 mole of silver carbonate, then 0.0376 moles of silver nitrate:

`=(1)/(2)* 0.0376 mol=0.0188 \text{mol of }Ag_2CO_3`

Moles of the silver carbonate formed = 0.0188 mol

Molar mass of silver carbonate = 275.7453 g/mol

Mass of silver carbonate :

`=275.7453 g/mol* 0.0188 mol=5.1840 g\approx 5.18 g`

The mass of silver carbonate precipitated is 5.18 grams.