Chlorine gas can be prepared according to the reaction: 16 HCl+2KMnO4->2MnCl2+8H2O When 25g HCl react with 25 g KMnO4, how many grams of Cl2 can be produced?
To find the grams of Cl2 produced, we first need to determine which reactant is limiting and then calculate the amount of Cl2 produced based on the balanced chemical equation.
First, let's calculate the moles of HCl and KMnO4 using their respective molar masses.
Molar mass of HCl: 1 g/mol (hydrogen) + 35.45 g/mol (chlorine) = 36.45 g/mol
Moles of HCl = mass of HCl / molar mass of HCl
= 25 g / 36.45 g/mol
≈ 0.686 mol
Molar mass of KMnO4: 39.10 g/mol (potassium) + 54.938 g/mol (manganese) + 4 * 16.00 g/mol (oxygen)
= 158.034 g/mol
Moles of KMnO4 = mass of KMnO4 / molar mass of KMnO4
= 25 g / 158.034 g/mol
≈ 0.158 mol
Next, let's determine the stoichiometry of the reaction. From the balanced chemical equation, we can see that the ratio of HCl to Cl2 is 16:1 and the ratio of KMnO4 to Cl2 is 2:1.
Since the stoichiometry is 16:1 for HCl:Cl2, we can say that for every 16 moles of HCl, 1 mole of Cl2 is produced.
However, we only have 0.686 moles of HCl, so using the stoichiometry, we can calculate the moles of Cl2 produced.
Moles of Cl2 = 0.686 mol HCl × (1 mol Cl2 / 16 mol HCl)
≈ 0.043 mol
Finally, we need to convert moles of Cl2 to grams using its molar mass.
Molar mass of Cl2: 2 * 35.45 g/mol (chlorine) = 70.90 g/mol
Grams of Cl2 = moles of Cl2 × molar mass of Cl2
≈ 0.043 mol × 70.90 g/mol
≈ 3.03 g
Therefore, when 25 g of HCl reacts with 25 g of KMnO4, approximately 3.03 grams of Cl2 will be produced.