Chemistry: Limiting Reactants

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

Write the correctly balanced equation:

Na₂CO₃ + CaCl₂*2H₂O ==> 2NaCl + CaCO₃

To find the limiting reactant, we will find moles of product produced by each reactant and compare them.

molar mass Na₂CO₃ = 106 g/mol

molar mass CaCl₂*2H₂O = 147 g/mol

molar mass CaCO₃ = 100 g/mol

molar mass NaCl = 58.4 g/mol

1) Finding limiting reactant:

1.20 g Na₂CO₃ x 1 mol/106 g x 1 mol CaCO₃/1 mol Na₂CO₃ x 100 g CaCO₃/mol = 1.13 g CaCO₃

1.9 g CaCl₂*2H₂O x 1 mol/147 g x 1 mol CaCO₃/1 mol CaCl₂*2H₂O x 100 g CaCO₃/mol = 1.29 g CaCO₃

So, Na₂CO₃ is LIMITING

2) Amount of CaCO₃ that can be made is 1.13 g CaCO₃ (see calculations in 1 above)

3) Amount of NaCl:

1.20 g Na₂CO₃ x 1 mol/106 g x 2 mol NaCl/mol Na₂CO₃ x 58.4 g NaCl/mol = 1.32 g NaCl