Chlorine gas can be prepared according to the reaction: 16HCI+2KMnO4->5CI2+2MnCI2+8H2O (b) how many grams of water will be produced when 75g of KCI are produced?
To determine how many grams of water will be produced when 75g of KCl are reacted, we need to use the stoichiometry of the balanced chemical equation.
According to the reaction: 16 HCl + 2 KMnO4 -> 5 Cl2 + 2 MnCl2 + 8 H2O
We can see that for every mole of KCl produced, 8 moles of water are produced. So, we need to calculate the amount of KCl in moles and then use the stoichiometric ratio to find the amount of water produced.
The molar mass of KCl is 39.10 g/mol.
Number of moles of KCl = mass / molar mass
Number of moles of KCl = 75 g / 39.10 g/mol
Number of moles of KCl = 1.92 mol
According to the stoichiometric ratio, 8 moles of water are produced for every 2 moles of KMnO4.
Therefore, the number of moles of water produced can be calculated using the ratio,
Number of moles of water = (1.92 mol KCl) x (8 mol H2O / 2 mol KMnO4)
Number of moles of water = 1.92 mol x 8 / 2
Number of moles of water = 7.68 mol
To find the mass of water produced, we can use the molar mass of water, which is 18.02 g/mol.
Mass of water = number of moles x molar mass
Mass of water = 7.68 mol x 18.02 g/mol
Mass of water = 138.34 g
Therefore, when 75 g of KCl are reacted, 138.34 grams of water will be produced.