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
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.
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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.
To find the mass of oxygen required to produce 100 g of KMnO4, we need to determine the stoichiometric coefficients of oxygen in both equations and connect the two equations through the intermediate compound, MnO2.
Let's start by determining the molar mass of KMnO4:
K (potassium) = 39.10 g/mol
Mn (manganese) = 54.94 g/mol
O (oxygen) = 16.00 g/mol
Total molar mass of KMnO4 = 39.10 + 54.94 + (4 × 16.00) = 158.03 g/mol
Now, let's use the first equation to calculate the moles of oxygen required to produce 2 moles of K2MnO4 (which is equivalent to 158.03 g):
From the equation: 2 MnO2 + 4 KOH + O2 → 2 K2MnO4 + 2 H2O
The coefficient of O2 is 1, so for 2 moles of K2MnO4, we need 2 moles of O2.
Next, we use the second equation to find the moles of oxygen required to produce 2 moles of KMnO4 (which is equivalent to 158.03 g):
From the equation: 3 K2MnO4 + 4 CO2 + 2 H2O → 2 KMnO4 + 4 KHCO3 + MnO2
The coefficient of KMnO4 is 2, so for 2 moles of KMnO4, we need 2 moles of O2.
Hence, we can conclude that for the production of 158.03 g of KMnO4, 2 moles of O2 are required.
To find the mass of oxygen required to produce 100 g of KMnO4, we can set up a proportion:
(2 moles O2 / 158.03 g KMnO4) = (x moles O2 / 100 g KMnO4)
x = (2 moles O2 / 158.03 g KMnO4) × (100 g KMnO4)
x ≈ 1.27 moles O2
Finally, we can calculate the mass of oxygen required:
Mass of oxygen = (1.27 moles O2) × (16.00 g/mol) = 20.32 g
Therefore, approximately 20.32 g of oxygen would need to be consumed to produce 100 g of KMnO4.
To find the mass of oxygen consumed in the reaction to produce 100 g of KMnO4, we need to use stoichiometry and the given balanced equations.
Let's break down the problem step by step:
Step 1: Convert the given mass of KMnO4 (100 g) to moles.
To do this, we need to know the molar mass of KMnO4. Let's find that first.
The molar mass of KMnO4 can be calculated as follows:
K (potassium) = 39.10 g/mol
Mn (manganese) = 54.94 g/mol
O (oxygen) = 16.00 g/mol x 4 = 64.00 g/mol
Molar mass of KMnO4 = 39.10 + 54.94 + 64.00 = 157.04 g/mol
Now, divide the given mass of KMnO4 by its molar mass to convert it to moles:
Number of moles of KMnO4 = 100 g / 157.04 g/mol
Step 2: Use stoichiometry to relate the moles of KMnO4 to the moles of O2.
From Equation 1, we know that 2 moles of MnO2 react with 1 mole of O2 to produce 2 moles of K2MnO4.
Based on Equation 1, we can set up the following ratio:
2 mol O2 / 2 mol K2MnO4
Step 3: Calculate the moles of O2 required.
Multiply the moles of KMnO4 (calculated in Step 1) by the ratio obtained in Step 2:
Number of moles of O2 = Number of moles of KMnO4 x (2 mol O2 / 2 mol K2MnO4)
Step 4: Convert the moles of O2 to grams.
To do this, we need to know the molar mass of O2, which is 32.00 g/mol.
Now, multiply the moles of O2 (calculated in Step 3) by its molar mass to convert it to grams:
Mass of O2 = Number of moles of O2 x 32.00 g/mol
This will give you the mass of oxygen required to produce 100 g of KMnO4.