How many grams of O2 are required to burn 46.0 grams of C5H12?
Wondering what "burn" means....
If you mean this reaction...
C5H12 + 8O2 >>>5CO2 + 6H2O
Then eight moles of O2 are required to combust one mole of Pentane. So how many moles of Pentane do you have in 46grams?
To find out how many grams of O2 are required to burn a given amount of a substance, we need to use the balanced chemical equation of the combustion reaction.
In the balanced equation, we can determine the ratio of the substance to oxygen (O2) required for complete combustion.
The balanced combustion equation for C5H12 is:
C5H12 + 8O2 → 5CO2 + 6H2O
Looking at the balanced equation, we see that for every 1 mole of C5H12, we need 8 moles of O2.
To solve the problem:
1. Convert the given grams of C5H12 to moles using its molar mass.
The molar mass of C5H12 = 5(12.01 g/mol) + 12(1.01 g/mol) = 72.15 g/mol
Moles of C5H12 = 46.0 g C5H12 / 72.15 g/mol = 0.636 mol C5H12
2. Use the mole ratio from the balanced equation to find the moles of O2 required.
From the balanced equation, the mole ratio of C5H12 to O2 is 1:8.
Moles of O2 = 0.636 mol C5H12 * (8 mol O2 / 1 mol C5H12) = 5.088 mol O2
3. Convert the moles of O2 to grams using its molar mass.
The molar mass of O2 = 2(16.00 g/mol) = 32.00 g/mol
Grams of O2 = 5.088 mol O2 * 32.00 g/mol = 163.02 g O2
So, 163.02 grams of O2 are required to burn 46.0 grams of C5H12.