Answer to Question #148129 in Chemistry for K

Question

as2s3+nano3+na2co3 -> na3aso4 + na2so4 + nano2 +co2 determine the theoretical mass of reactants needed to form 50 g of sodium arsenate

Answer 1

As₂S₃+14NaNO₃+6Na₂CO₃ -> 2Na₃AsO₄ + 3Na₂SO₄ + 14NaNO₂ +6CO₂

M(Na₃AsO₄)=208 g/mol

n (Na₃AsO₄)=m/M = 50/208 = 0.24 mol

M (As₂S₃) = 246 g/mol

M (NaNO₃) = 85 g/mol

M (Na₂CO₃) = 106 g/mol

n (As₂S₃) = 1/2 x n (Na₃AsO₄) = 1/2 x 0.24 = 0.12 mol

m (As₂S) = n x M = 0.12 x 246 = 29.52 g

n (NaNO₃) = 14/2 x n (Na₃AsO₄) = 14/2 x 0.24 = 1.68 mol

m (NaNO₃) = n x M = 1.68 x 85 = 142.8 g

n (Na₂CO₃) = 6/2 x n (Na₃AsO₄) = 6/2 x 0.24 = 0.72 mol

m (Na₂CO₃) = n x M = 0.72 x 106 = 76.32 g