Question

If 0.220 mol of solid CaCO3 and 670 mL of 0.433 M aqueous H2SO4 are reacted stoichiometrically according to the balanced equation. How many grams of solid CaSO4 are produced?
H2SO4(aq) + CaCO3(s) → CO2(g) + CaSO4(s) + H2O(l)

Answer 1

Answer:

29.92 grams of solid calcium sulfate are produced.

Explanation:

$H_2SO_4(aq) + CaCO_3(s)\rightarrow CO_2(g) + CaSO_4(s) + H_2O(l)$

670 mL of 0.433 M aqueous sulfuric acid:

Moles of sulfuric acid = n

Volume of the sulfuric acid = V = 670 mL = 0.670 L ( 1mL = 0.001 L)

Molarity of the sulfuric acid = M = 0.433 M

$M=\frac{n}{V(L)}$

$n=M\times V=0.433 M\times 0.670 L=0.29011mol$

Moles of sulfuric acid = 0.29011 mol

Moles of calcium carbonate = 0.220 mol

According to reaction, 1 mole of calcium carbonate reacts with 1mole of sulfuric acid.

Then 0.220 moles of calcium carbonate will reacts with :

$\frac{1}{1}\times 0.220 mol=0.220 mol$ sulfuric acid.

This means that calcium carbonate is in limiting amount and mass of calcium sulfate will depend upon it.

According to reaction, 1 mole of calcium carbonate gives with 1 mole of calcium sulfate .

Then 0.220 moles of calcium carbonate will give with :

$\frac{1}{1}\times 0.220 mol=0.220 mol$ calcium sulfate

Mass of 0.220 moles of calcium sulfate

0.220 mol × 136 g/mol = 29.92 g

29.92 grams of solid calcium sulfate are produced.