Long way that requires no new learning. Just make it two stoichiometry. problems. Use equation 2 to calculate grams needed of K2MnO4 to make your 100 g KMnO4. Then switch to equation 1 and calculate grams O2 need to make that many grams of K2MnO4
Short way (which may not be much shorter):
You want 100 g KMnO4. That is 100/158 = approx 0.6 mol. So you want the two equations to provide that . Therefore, we need to know mols O2 for mols KMnO4.
.
2 MnO2 + 4 KOH + O2 → 2 K2MnO4 + 2 H2O
3 K2MnO4 + 4 CO2 + 2 H2O → 2 KMnO4 + 4 KHCO3 + MnO2
From eqn 1 we have: 1 mol O2 = 2 mols K2MnO4
from eqn 2 we have: 3 mols K2MnO4 = 2 mol KMnO4
If we make mols K2MnO4 in the two equation equal (multiply new eqn 1 by 3 and new eqn 2 by 2 to get this:
3 mol O2 = 6 mols K2MnO4 and new eqn 2 multiplied by 2 is
6 mols K2MnO4 = 4 mol KMnO4
Voila!, Since both equations = same number of mols K2MnO4, that means that 3 mols O2 = 4 mols KMnO4. So if I want 0.6 mols KMnO4 I must have
0.6 mols KMnO4 x (3 mols O2/4 mols KMnO4) = ? mols O2. Convert that to grams.
Consider the following two equations used in the preparation of KMnO4:
2 MnO2 + 4 KOH + O2 → 2 K2MnO4 + 2 H2O
3 K2MnO4 + 4 CO2 + 2 H2O → 2 KMnO4 + 4 KHCO3 + MnO2
What mass of oxygen must be consumed in order to make 100 g of KMnO4.
I don't even know how to link these two equations. Any help is much appreciated
1 answer