Question

what is the mass of impurity in a sample of CaCO3 if 0.040mole of CO2 was formed when 5.0g of the sample reacted with excessbdilute HCL

Answer 1

CaCO₃ + 2HCl → CaCl₂ + CO₂ + H₂O

n(CO₂) = n(CaCO₃) = 0.04 mol

M(CaCO₃) = 100.08 g/mol

$m = n \times M$

m(CaCO₃) $= 0.04 \times 100.08 = 4.00 \;g$

Δm = 5.0 – 4.0 = 1.0 g

Answer: 1.0 g