Calculate the mass of water formed by the interaction of 280 g of potassium hydroxide with a sulfuric acid solution.

2KOH + H2SO4 = K2SO4 + 2H2O.

To solve the problem, we need molar masses: Mr (KOH) = 56 g / mol, Mr (H2O) = 18 g / mol. The reaction shows that 1 mole of water is formed from 1 mole of hydroxide. That is, from 56 g of hydroxide, 18 g of water is obtained. Let’s make the proportion:

56 g – 18 g

280 g – x g.

Hence x = 280 * 18/56 = 90 g – the desired mass of water.