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

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