Answer:
HCl is the limiting reactant.
20.2 grams = theoretical yield Cl2
actual yield = 16.75 grams Cl2
Explanation:
Step 1: Data given
Mass of MnO2 = 43.5 grams
Molar mass MnO2 = 86.94 g/mol
Mass of HCl = 41.5 grams
Molar mass HCl = 36.46 g/mol
Step 2: The balanced equation
4HCl(aq)+MnO2(s)⟶MnCl2(aq)+2H2O(l)+Cl2(g)
Step 3: Calculate moles
Moles = mass / molar mass
Moles MnO2 = 43.5 grams / 86.94 g/mol
Moles MnO2 = 0.500 moles
Moles HCl = 41.5 grams / 36.46 g/mol
Moles HCl = 1.14 moles
Step 4: Calculate the limiting reactant
For 4 moles HCl we need 1 mol MnO2 to produce 1 mol MnCl2, 2 moles H2O and 1 mol Cl2
HCl is the limiting reactant. I will completely be consumed (1.14 moles).
MnO2 is in excess. There will react 1.14 /4 = 0.285 moles
There will remain 0.500 - 0.285 = 0.215 moles MnO2
Step 5: Calculate moles Cl2
For 4 moles HCl we need 1 mol MnO2 to produce 1 mol MnCl2, 2 moles H2O and 1 mol Cl2
For 1.14 moles HCl we'll have 1.14/4 = 0.285 moles Cl2
Step 6: Calculate mass Cl2
Mass Cl2 = moles Cl2 * molar mass Cl2
Mass Cl2 = 0.285 moles * 70.9 g/mol
Mass Cl2 = 20.2 grams = theoretical yield
Step 7: Calculate actual yield
% yield = (actual yield / theoretical yield) *100%
0.829 = actual yield / 20.2 grams
actual yield = 0.829 * 20.2 grams
actual yield = 16.75 grams Cl2
