Question

how many grams of water can be produced when 65.5 grams of sodium hydroxide react with excess sulfuric acid.

Answer 1

2NaOH+H2SO4--> Na2SO4+2H2O

80g NaOH(40g/mol)--> 36g H2O (18g/mol)

65.5 x (1/40g/mol)= 1.64 moles of NaOH

1.64 x 18g/mol= 29.5 g H2O