the combustion of a sample of butane, C4H10 (lighter fluid) produced 3.75 moles of water.
how to solve for: (with solution)
1.) how many grams of butane burned?
2.) How much oxygen was used up in grams?
(1) the reaction equation will tell you how many moles of butane to use -- convert to grams
(2) again, consult the equation and convert moles to grams
To solve these problems, we need to follow a step-by-step process. We'll start with the balanced chemical equation for the combustion of butane, C4H10:
2C4H10 + 13O2 → 8CO2 + 10H2O
1.) How many grams of butane burned?
We are given the number of moles of water produced (3.75 moles). From the balanced equation, we can see that for every 2 moles of butane burned, we produce 10 moles of water. This means that the mole ratio between water and butane is 10/2 = 5.
To find the number of moles of butane burned, we can use the mole ratio:
Moles of butane = Moles of water x (Mole ratio of butane to water)
= 3.75 moles x 5
= 18.75 moles
Now we need to convert moles of butane to grams. To do this, we need to know the molar mass of butane. The molar mass of C4H10 is:
(4 x atomic mass of carbon) + (10 x atomic mass of hydrogen)
= (4 x 12.01 g/mol) + (10 x 1.01 g/mol)
= 48.04 g/mol + 10.1 g/mol
= 58.14 g/mol
Finally, we can calculate the mass of butane burned:
Mass of butane = Moles of butane x Molar mass of butane
= 18.75 moles x 58.14 g/mol
= 1090.125 g
≈ 1090.13 g
Therefore, approximately 1090.13 grams of butane were burned.
2.) How much oxygen was used up in grams?
From the balanced equation, we can see that for every 13 moles of oxygen used, we produce 10 moles of water. This means that the mole ratio between oxygen and water is 10/13.
To find the number of moles of oxygen used, we can use the mole ratio:
Moles of oxygen = Moles of water x (Mole ratio of oxygen to water)
= 3.75 moles x (10/13)
= 2.884 moles
Now we need to convert moles of oxygen to grams. To do this, we need to know the molar mass of oxygen, which is approximately 16.00 g/mol.
Finally, we can calculate the mass of oxygen used:
Mass of oxygen = Moles of oxygen x Molar mass of oxygen
= 2.884 moles x 16.00 g/mol
= 46.144 g
≈ 46.14 g
Therefore, approximately 46.14 grams of oxygen were used up in the combustion of butane.