Chlorine gas can be prepared according to the reaction: 16 HCl+2KMnO4->2MnCl2+8H2O how many moles of MnCl2 can be produced when 25g KMnO4 are mixed with 85g HCl?
To determine how many moles of MnCl2 can be produced, we need to calculate the moles of KMnO4 and HCl and then use the stoichiometry of the reaction.
First, let's calculate the moles of KMnO4:
Molar mass of KMnO4 = 39.1 g/mol (K) + 54.9 g/mol (Mn) + 4(16 g/mol) (O) = 158.0 g/mol
Moles of KMnO4 = Mass of KMnO4 / Molar mass of KMnO4
Moles of KMnO4 = 25 g / 158.0 g/mol
Moles of KMnO4 ≈ 0.1587 mol
Next, let's calculate the moles of HCl:
Molar mass of HCl = 1 g/mol (H) + 35.5 g/mol (Cl) = 36.5 g/mol
Moles of HCl = Mass of HCl / Molar mass of HCl
Moles of HCl = 85 g / 36.5 g/mol
Moles of HCl ≈ 2.33 mol
Now, let's use the stoichiometry of the reaction to convert the moles of KMnO4 to moles of MnCl2:
From the balanced chemical equation, we can see that the ratio of KMnO4 to MnCl2 is 2:2 or 1:1.
Moles of MnCl2 = Moles of KMnO4 ≈ 0.1587 mol
Therefore, approximately 0.1587 mol of MnCl2 can be produced when 25g of KMnO4 are mixed with 85g of HCl.