Answer to Question #164694 in Chemistry for Brit Lessy

Question

1.20 grams of sodium carbonate reacts with 1.9 grams of calcium chloride dihydrate to form sodium chloride and calcium carbonate in a double replacement type reaction.

1) calculate with reactant is the limiting reactant. Show the work and label which reactant is limiting.

2) once you know which is the limiting reactant, calculate the amount of calcium carbonate which can be made.

3) once you know the limiting reactant, calculate the amount of sodium chloride which can be made.

Show work for every question

Answer 1

Na₂CO₃+CaCl₂ = CaCO₃+2NaCl

M (Na₂CO₃) = 105.99 g/mol

M (CaCl₂) = 110.98 g/mol

n (Na₂CO₃) = m/M = 1.20/105.99 = 0.011 mol

n (CaCl₂) = 1.9/110.98 = 0.017 mol

Na₂CO₃ is the limiting reactant.

n (CaCO₃) = n (Na₂CO₃) = 0.011 mol

M (CaCO₃) = 100.09 g/mol

m (CaCO₃) = n x M = 0.011 x 100.09 = 1.1 g

n (NaCl) = 2 x n (Na₂CO₃) = 0.011 x 2 = 0.022 mol

M (NaCl) = 58.44 g/mol

m (NaCl) = n x M = 0.022 x 58.44 = 1.3 g