Answer:
31.44 g KMnO4
Explanation:
M(MnCl2) = M(Mn) + 2M(Cl) = 54.9 + 2*35.5 =125.9 g/mol
25.00 g /125.9 g/mol =0.1990 mol MnCl2
M(PbO2) = M(Pb) + 2M(O2) = 207.2 +2*16.0 =239.2 g/mol
100g/139.2 g/mol = 0.7184 mol PbO2
2 KCl + 2 MnCl2 + 5 PbO2 + 4 HCl → 2 KMnO4 + 5 PbC12 + 2 H2O
from reaction 2 mol 5 mol
given 0.1990mol 0.7184 mol
for 2 mol MnCl2 ----- 5 mol PbO2
for 0.1990 mol MnCl2 ---- x mol PbO2
x = (0.1990 *5)/2 = 0.4975 mol PbO2
So, for 0.1990 MnCl2 we need 0.4975 mol PbO2, but we have 0.7184 mol PbO2. That means that we have excess of PbO2, and we are going to use for further calculation 0.1990 mol MnCl2
2 KCl + 2 MnCl2 + 5 PbO2+ 4 HCl → 2 KMnO4 + 5 PbC12 + 2 H2O
from reaction 2 mol 2 mol
given 0.1990 mol
gotten 0.1990 mol
We got 0.1990 mol KMnO4.
M(HMnO4) = M(K) + M(Mn) +4M(O) = 158.0 g/mol
m(KMnO4) = 0.1990 mol*158.0 g/mol = 31.44 g
