Answer:
The limiting reactant is Hg2Cl2.
Explanation:
Step 1: Data given
Mass of each reactant = 5.0 grams
KMnO4 MM=158 g/mol
Hg2Cl2 MM=472.1 g/mol
HCl MM=36.5 g/mol
HgCl2 MM=271.5 g/mol
MnCl2 MM=125.8 g/mol
KCl MM=74.6 g/mol
H2O MM=18 g/mol)
Step 2: The balanced equation
2KMnO4 + 5Hg2Cl2 + 16HCl -> 10HgCl2 + 2MnCl2 + 2KCl + 8H2O
Step 3: Calculate moles
KMnO4 = 5.00 grams / 158 g/mol = 0.0316 mol
Hg2CL2 = 5.00 grams / 472.1 g/mol = 0.0106 mol
HCl = 5.00 grams / 36.5 g/mol = 0.137 mol
Step 3: Calculate limiting reactant
For 2 moles of KMno4 we need 5 moles of Hg2Cl2 and 16 moles of HCl
Hg2Cl2 has the smallest amount of moles.
For 5 moles Hg2Cl2 ( 0.0106 mol) we need 0.0106 / (5/2) = 0.00424 mol KMnO4
For 5 moles Hg2Cl2 we need (16/5) *0.0106 = 0.03392 moles of HCl
So the limiting reactant is Hg2Cl2.
Step 4: Calculate moles of product produced:
2*0.0106 = 0.0212 moles of HgCl2
(2/5) * 0.0106 = 0.00424 moles of MnCl2 and 0.00424 moles of KCl
(8/5) * 0.0106 = 0.01696 moles H2O
