Question

In the reaction below 25.00 g of MnC1, is reacted with 100.0 g of PbO2, excess KCl, and excess

HCI.
2 KCl (aq) + 2 MnCl2 (aq) + 5 PbO2 (6) + 4 HC1 (aq) → 2 KMnO4 (aq) + 5 PbC12 (6) + 2 H2O()(EQ 10.3)
a. How many grams of KMnO4 can be produced by this reaction? Use an ICE table.​

Answer 1

31.44 g KMnO4

Step-by-step 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