Question

What is the mass of sodium hydroxide needed to produce 2.7 x 10⁵ grams of liquid water? 

Answer 1

$NaOH_{(aq)} + HA_{(aq)} \to NaA_{(aq) } + H_2O_{(l)}$

1 mole of NaOH produces 1 mole of water

$\therefore \dfrac{2.7 × 10^5 g}{40g/mol}$ of NaOH produces 6750 moles of water

6750 moles of water = 121500g of water produced