Chlorine gas can be prepared according to the reaction: 16 HCl+2KMnO4->2MnCl2+8H2O how many grams of water will be produced when 75g of KCl are produced?

Answer 1

To solve this problem, we need to first determine the stoichiometry of the reaction. From the balanced equation, we can see that 2 moles of KMnO4 react with 16 moles of HCl to form 8 moles of water.

1 mole of KMnO4 has a molar mass of 158.04 g/mol, so 75 g of KMnO4 is equal to (75 g) / (158.04 g/mol) = 0.475 mol of KMnO4.

According to the reaction stoichiometry, 2 moles of KMnO4 produce 8 moles of water, so 0.475 mol of KMnO4 will produce (8 moles of water) × (0.475 mol of KMnO4 / 2 moles of KMnO4) = 1.9 moles of water.

The molar mass of water is 18.02 g/mol, so 1.9 moles of water is equal to (1.9 moles) × (18.02 g/mol) = 34.23 g of water.

Therefore, when 75 g of KMnO4 react, 34.23 g of water will be produced.