Answer:
29.92 grams of solid calcium sulfate are produced.
Explanation:
[tex]H_2SO_4(aq) + CaCO_3(s)\rightarrow CO_2(g) + CaSO_4(s) + H_2O(l)[/tex]
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
[tex]M=\frac{n}{V(L)}[/tex]
[tex]n=M\times V=0.433 M\times 0.670 L=0.29011mol[/tex]
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 :
[tex]\frac{1}{1}\times 0.220 mol=0.220 mol[/tex] 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 :
[tex]\frac{1}{1}\times 0.220 mol=0.220 mol[/tex] 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.
