What is the mass of sodium hydroxide needed to produce 2.7 x 10⁵ grams of liquid water?
"NaOH_{(aq)} + HA_{(aq)} \\to NaA_{(aq) } + H_2O_{(l)}"
1 mole of NaOH produces 1 mole of water
"\\therefore \\dfrac{2.7 \u00d7 10^5 g}{40g\/mol}" of NaOH produces 6750 moles of water
6750 moles of water = 121500g of water produced
Comments
Leave a comment