Question

If 50mL of 0.15M calcium sulfide is added to 30mL of 0.35M ammonium carbonate, how many grams of precipitate will form

Answer 1

0.750 g

Step-by-step explanation:

The reaction that takes place is:

• CaS (aq) + (NH₄)₂CO₃ (aq) → CaCO₃ (s) + (NH₄)₂S (aq)

First we calculate how many moles of each reactant were added, using the given volumes and concentrations:

• CaS ⇒ 50 mL * 0.15 M = 7.5 mmol CaS

• (NH₄)₂CO₃ ⇒ 30 mL * 0.35 M = 10.5 mmol (NH₄)₂CO₃

Given that they react in a 1:1 ratio, and that there are less CaS moles than (NH₄)₂CO₃ moles, CaS is the limiting reactant.

• 7.5 mmoles of CaS will produce 7.5 mmoles of CaCO₃.

We now convert 7.5 mmol CaCO₃ into mg, using its molar mass:

• 7.5 mmol * 100 mg/mmol = 750 mg

Finally we convert 750 mg to g:

• 750 mg / 1000 = 0.750 g