How many grams of Ag2CO3 will precipitate when excess (NH4)2CO3 solution is added to 74.0 mL of 0.532 M AgNO3 solution?

Question

2AgNO₃(aq) + (NH₄)₂CO₃(aq) Ag₂CO₃(s) + 2NH₄NO₃(aq)

Answer 1

First you find the moles of AgNO₃. You do this by multiplying the mole conc. by the volume.

74.0/1000 * 0.532M =0.03937 moles of AgNO₃.

You then find the moles of (NH₄)CO₃ that will be used to produce Ag₂CO₃.

The mole ratio of AgNO₃ and (NH₄)CO₃ is 2:1 so 0.03937 moles of AgNO₃ will react with 0.01969 moles of (NH₄)₂CO_3.

Because the 0,01969 moles of Carbonate is present 0.01969 moles of Ag₂CO₃ will be produced.

Multiply the moles of Ag₂CO₃ by its RMM:

0.01969*275.74=5.43 grams of Ag₂CO_3.