Question

compute the mass of CaSO4 that can be prepared by the reaction of 3.2900g of H2SO4 with 3.1660g of CaCO3

Answer 1

4.3065 g First, lookup the atomic weights of all the elements involved. Atomic weight of Calcium = 40.078 Atomic weight of Carbon = 12.0107 Atomic weight of Hydrogen = 1.00794 Atomic weight of Oxygen = 15.999 Atomic weight of Sulfur = 32.065 Now calculate the molar masses of the reactants and product Molar mass H2SO4 = 2 * 1.00794 + 32.065 + 4 * 15.999 = 98.07688 g/mol Molar mass CaCO3 = 40.078 + 12.0107 + 3 * 15.999 = 100.0857 g/mol Molar mass CaSO4 = 40.078 + 32.065 + 4 * 15.999 = 136.139 g/mol The balanced reaction for H2SO4 with CaCO3 is CaCO3 + H2SO4 ==> CaSO4 + H2O + CO2 So it takes 1 mole each of CaCO3 and H2SO4 to produce 1 mole of CaSO4. Let's see how many moles of CaCO3 and H2SO4 we have. CaCO3: 3.1660 g / 100.0857 g/mol = 0.031632891 mol H2SO4: 3.2900g / 98.07688 g/mol = 0.033545113 mol We have a slight excess of H2SO4, so the amount of CaCO3 is the limiting reactant and we should have 0.031632891 moles of product. To determine its mass, multiply the number of moles by the molar mass computed earlier. 0.031632891 mol * 136.139 g/mol = 4.306470148 g Since we have 5 significant figures in our data, round the result to 5 figures, giving 4.3065 g

Answer 2

Answer: 4.3020 grams of calcium sulfate will be produced.

Explanation:

To calculate the number of moles, we use the formula:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$    ....(1)

• For calcium carbonate:

Given mass of calcium carbonate = 3.1660 g

Molar mass of calcium carbonate = 100.0869 g/mol

Putting values in above equation, we get:

$\text{Moles of calcium carbonate}=\frac{3.1660g}{100.0869g/mol}=0.0316mol$

• For sulfuric acid:

Given mass of sulfuric acid = 3.2900 g

Molar mass of sulfuric acid = 98.079 g/mol

Putting values in above equation, we get:

$\text{Moles of sulfuric acid}=\frac{3.2900g}{98.079g/mol}=0.0335mol$

For the given reaction:

$CaCO_3+H_2SO_4\rightarrow CaSO_4+H_2CO_3$

By Stoichiometry of the reaction:

1 mole of calcium carbonate reacts with 1 mole of sulfuric acid

So, 0.0316 moles of calcium carbonate will react with = $\frac{1}{1}\times 0.0316mol=0.0316mol$

As, the given amount of sulfuric acid is more than the required amount. Hence, it is present in excess and is considered as an excess reagent.

Calcium carbonate is considered as a limiting reagent because it limits the formation of product.

• By Stoichiometry of the reaction:

1 mole of calcium carbonate produces 1 mole of calcium sulfate

So, 0.0316 moles of calcium carbonate will produce = $\frac{1}{1}\times 0.0316mol=0.0316mol$ of calcium sulfate.

Now, to calculate the mass of calcium sulfate, we use equation 1:

Molar mass of calcium sulfate = 136.14 g/mol

Moles of calcium sulfate = 0.0316mol

Putting values in equation 1, we get:

$0.0316mol=\frac{Mass of calcium sulfate}}{136.14g/mol}\\\\\text{Mass of calcium sulfate}=4.3020g$

Hence, 4.3020 grams of calcium sulfate will be produced.