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?
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.