Answer to Question #98977 in General Chemistry for heje

Question

A 43.2 sample of CaCO3 was treated with an excess of aqueous H2SO4, producing calcium sulfate (CaSO4), 3.65 g of water and CO2(g). What was the % yield of H2O?

Answer 1

According to the reaction:

CaCO₃ + H₂SO₄ = CaSO₄ + H₂O + CO₂

From here, the theoretical yield of water equals::

m_t (H₂O) = m(CaCO₃) × Mr(H₂O) / Mr(CaCO₃) = 43.2 g × 16 g/mol / 100 g/mol = 6.912 g

As the actual (m_a) yield of water equals 3.65 g, the percent yield is:

% yield = (m_a / m_t) × 100% = (3.65 g / 6.912 g) × 100% = 52.81%

Answer: 52.81%