The following reaction shows sodium carbonate reacting with calcium hydroxide.

Na2CO3 + Ca (OH) 2 → NaOH + CaCO3 How many grams of NaOH are produced from 20.0 grams of Na2CO3? (Molar mass of Na = 22.989 g/mol, C = 12.01 g/mol, O = 15.999 g/mol, Ca = 40.078 g/mol, H = 1.008 g/mol)

12.2 grams

15.1 grams

24.4 grams

30.2 grams

Ans: 15.1 grams

Given reaction:

Na2CO3 + Ca (OH) 2 → 2NaOH + CaCO3

Mass of Na2CO3 = 20.0 g

Molar mass of Na2CO3 = 105.985 g/mol

# moles of Na2CO3 = 20/105.985 = 0.1887 moles

Based on the reaction stoichiometry: 1 mole of Na2CO3 produces 2 moles of NaOH

# moles of NaOH produced = 0.1887*2 = 0.3774 moles

Molar mass of NaOH = 22.989 + 15.999 + 1.008 = 39.996 g/mol

Mass of NaOH produced = 0.3774*39.996 = 15.09 grams