Qualitative analysis

Question

12.5g of ZnCO3 were heated very strongly to a constant mass and the residue treated with excess HCl. Calculate the mass of ZnCl2 that would be obtained 

Answer 1

Start by writing the balanced reaction out.

ZnCO₃ + 2HCl --> ZnCl₂ + H₂O + CO₂

 

Excess HCl means you have plenty of it so the limiting reagent is ZnCO₃

Convert 12.5g ZnCO₃ to moles ZnCO₃

 

Molar mass ZnCO₃ = 65.39g/mol + 12.01g/mol + 3*16g/mol = 125.4g/mol ZnCO₃

12.5g ZnCO₃ * 1 mol ZnCO₃ / 125.4g = 0.1 mol ZnCO₃

 

The stoichiometry is 1:1 ZnCO₃ to ZnCl₂ so it's the same mole

 

Multiply the mole ZnCl₂ by it's molar mass to obtain the mass in grams

Molar mass ZnCl₂ = 65.39g/mol + 2*35.45g/mol = 136.28g/mol ZnCl₂

 

0.1 mol ZnCl₂ * 136.28g/mol ZnCl₂ = 13.63g ZnCl₂