Answer:
53.9%
Explanation:
1 mole of BaCO₃ yields 1 mole of CO₂,
1 mole of SrCO₃ yields 1 mole of CO₂
m₁ = mass of BaCO₃
m₂ = mass SrCO₃
molar mass of SrCO₃ = 147.63 g/mol
molar mass of BaCO₃ = 197.34 g/mol
molar mass of CO₂ = 44.01 g/mol
mole of CO₂ in 0.211 g = 0.211 g / 44.01 = 0.00479
mole of BaCO₃ = m₁ / 197.34
mole of SrCO₃ = m₂ / 147.63
mole of BaCO₃ + mole of SrCO₃ = 0.00479
(m₁ / 197.34) + (m₂ / 147.63) = 0.00479
147.63 m₁ + 197.34 m₂ = 139.55
m₁ + m₂ = 0.8
m₁ = 0.8 - m₂
147.63 (0.8 - m₂) + 197.34 m₂ = 139.55
118.104 - 147.63 m₂ + 197.34 m₂ = 139.55
49.71 m₂ = 139.55 - 118.104 = 21.446
m₂ = 21.446 / 49.71 = 0.431
the percentage of m₂ ( SrCO₃ ) in the sample = 0.431 / 0.8 = 0.539 × 100 = 53.9%
