To calculate the grams of O2 required for the combustion of 25.9 g of ethylcyclopentane, we need to first write the balanced chemical equation for the combustion of ethylcyclopentane:
C11H20 + 15O2 → 11CO2 + 10H2O
From the balanced equation, we can see that 15 moles of O2 are required to react with 1 mole of C11H20.
To calculate the grams of O2 required, we can use the following steps:
1. Calculate the number of moles of C11H20:
moles of C11H20 = mass of C11H20 / molar mass of C11H20
moles of C11H20 = 25.9 g / 140.28 g/mol
moles of C11H20 = 0.1847 mol
2. Calculate the number of moles of O2 required:
moles of O2 = 15 moles of O2 / 1 mole of C11H20 * 0.1847 moles of C11H20
moles of O2 = 2.7705 moles
3. Calculate the mass of O2 required:
mass of O2 = moles of O2 * molar mass of O2
mass of O2 = 2.7705 mol * 32.00 g/mol
mass of O2 = 88.97 g
Therefore, the grams of O2 required for the combustion of 25.9 g of ethylcyclopentane is 88.97 g.