Question

what mass, in grams, of zinc sulfate heptahydrate would she need to use to generate the 1.000 kg of caso₄? molar mass caso₄ = 136.14 g/mol molar mass znso₄·7h₂o = 287.54 g/mol znso₄·7h₂o + cacl₂·2h₂o → caso₄(s) + zncl₂ + 9h₂o ? g znso₄·7h₂o

Explanation:

Step1: Convert mass of \( \text{CaSO}_4 \) to moles

Given mass of \( \text{CaSO}_4 = 1.000 \, \text{kg} = 1000 \, \text{g} \), molar mass of \( \text{CaSO}_4 = 136.14 \, \frac{\text{g}}{\text{mol}} \)
Moles of \( \text{CaSO}_4 = \frac{\text{mass}}{\text{molar mass}} = \frac{1000 \, \text{g}}{136.14 \, \frac{\text{g}}{\text{mol}}} \approx 7.345 \, \text{mol} \)

Step2: Determine moles of \( \text{ZnSO}_4 \cdot 7\text{H}_2\text{O} \) from stoichiometry

From the balanced equation, the mole ratio of \( \text{ZnSO}_4 \cdot 7\text{H}_2\text{O} \) to \( \text{CaSO}_4 \) is \( 1:1 \). So moles of \( \text{ZnSO}_4 \cdot 7\text{H}_2\text{O} = \) moles of \( \text{CaSO}_4 = 7.345 \, \text{mol} \)

Step3: Calculate mass of \( \text{ZnSO}_4 \cdot 7\text{H}_2\text{O} \)

Molar mass of \( \text{ZnSO}_4 \cdot 7\text{H}_2\text{O} = 287.54 \, \frac{\text{g}}{\text{mol}} \)
Mass of \( \text{ZnSO}_4 \cdot 7\text{H}_2\text{O} = \text{moles} \times \text{molar mass} = 7.345 \, \text{mol} \times 287.54 \, \frac{\text{g}}{\text{mol}} \approx 2112 \, \text{g} \)

Answer:

\( 2112 \) (or more precisely, after detailed calculation: \( \frac{1000}{136.14} \times 287.54 \approx 2112 \) g)