Question

Consider the reaction MnO2 + 4HCI → MnCl2 + Cl2 + 2H2O. If the both the

reactants have a mass of 100 grams, which of the two will be used up first? How

many grams of MnCl2 will be produced after the reaction? (MnO2 =

86.94 g/mol, HC1 = 36.46 g/mol, MnCl2 = 125.8 g/mol)

Answer 1

Calculate $\nu:$

$\nu(MnO_2) =m(MnO_2)/M(MnO_2)=100g/(86.94 g/mol)=1.15 mol$

$\nu(HCl) =m(HCl)/M(HCl)=100g/(36.46 g/mol)=2.74 mol$

1 mol of $MnO_2$ reacts with 4 mol of $HCl$ , 1.15 mol of $MnO_2$ reacts with 4.6 mol of $HCl$

$MnO_2$ in excess

HCl will be used up first - it limiting reagent.

According to the equation we get 1/4 mol of $MnCl_2$ from 1 mol of HCl, therefore:

$m(MnCl_2)=\nu(MnCl_2)\cdot M(MnCl_2)=1/4\cdot \nu(HCl)\cdot M(MnCl_2)=$

$=0.25\cdot 2.74 mol\cdot 125.8 g/mol=86.17g$