A sample that contains only SrCO3 and BaCO3 weighs 0.800 g. When it is dissolved in excess acid, 0.211 g car - bon dioxide is liberated. What percentage of SrCO3 did the sample contain?

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%