Explanation:
Equation of the reaction:
2KMnO4(aq) + 16HCl(aq) --> 2MnCl2(aq) + 2KCl(aq) + 5Cl2(g) + 8H2O(aq)
To calculate the limiting reagent, we need to calculate the number of moles of the reactants :
KMnO4:
Molar mass = (39 + 55 + (16*4))
= 158 g/mol
Number of moles = mass/molar mass
= 229.19/158
= 1.4506 mol
HCl:
Molar mass = 1 + 35.5
= 36.5 g/mol
Number of moles = 526.64/36.5
= 14.428 mol
By stoichiometry, 2 moles of KMnO4 reacted with 16 moles of HCl
The limiting reagent :
14.428 moles of HCl * 2 moles of KMnO4/16moles of HCl
= 1.8035 moles of KMnO4 is required to react with 14.428 moles of HCl
But there's 1.4506 moles of KMnO4. Therefore, KMnO4 is the limiting reagent.
Mass of the products:
KCl:
2 moles of KMnO4 will produce 2 moles of KCl
Moles of KCl = 1 * 1.4506 mol
= 1.4506 mol
Molar mass = 39 + 35.5 = 74.5 g/mol
Mass of KCl = 74.5 * 1.4506
= 108.07 g
MnCl2:
2 moles of KMnO4 will produce 2 moles of MnCl2
Number of moles of MnCl2 = 1 * 1.4506
= 1.4506 mol
Molar mass = 55 + (35.5*2)
= 126 g/mol
Mass of MnCl2= 1.4506 * 126
= 182.78 g
Cl2:
2 moles of KMnO4 will produce 5 moles of Cl2
Number of moles of Cl2 = 5/2 * 1.4506
= 3.6265 mol
Molar mass of Cl2 = 35.5*2
= 71 g/mol
Mass of Cl2 = 71* 3.6265
= 257.4815 g
H2O:
2 moles of KMnO4 will produce 8 moles of H2O
Number of moles of H2O = 8/2 * 1.4506
= 5.80 mol
Molar mass of H2O =( 1*2) + 16
= 18 g/mol
Mass of H2O = 18*5.80
= 104.44 g
