How many grams of H2O (Mm = 18.0 g/mol) are produced in the combustion of 70.0 g of C4H10 (Mm = 58.1 g/mol)?
2C4H10 + 13O2 8CO2 + 10H2O
To find the number of grams of H2O produced in the combustion of C4H10, you need to use stoichiometry.
The balanced equation for the combustion of C4H10 is:
2C4H10 + 13O2 -> 8CO2 + 10H2O
From the balanced equation, you can see that for every 2 moles of C4H10 burned, 10 moles of H2O are produced.
Step 1: Calculate the number of moles of C4H10
Given that the molar mass of C4H10 is 58.1 g/mol and you have 70.0 g of C4H10:
Moles of C4H10 = Mass of C4H10 / Molar mass of C4H10
Moles of C4H10 = 70.0 g / 58.1 g/mol
Step 2: Use the mole ratio from the balanced equation to find moles of H2O
From the balanced equation, you know that the mole ratio of C4H10 to H2O is 2:10.
Moles of H2O = Moles of C4H10 * (10 moles of H2O / 2 moles of C4H10)
Step 3: Convert moles of H2O to grams
To convert moles to grams, use the molar mass of H2O, which is 18.0 g/mol.
Grams of H2O = Moles of H2O * Molar mass of H2O
Calculating these steps:
Step 1: Moles of C4H10
Moles of C4H10 = 70.0 g / 58.1 g/mol
Moles of C4H10 ≈ 1.2047 mol
Step 2: Moles of H2O
Moles of H2O = 1.2047 mol * (10 mol H2O / 2 mol C4H10)
Moles of H2O ≈ 6.0235 mol
Step 3: Grams of H2O
Grams of H2O = 6.0235 mol * 18.0 g/mol
Grams of H2O ≈ 108.4 g
Therefore, approximately 108.4 grams of H2O are produced in the combustion of 70.0 grams of C4H10.
mols C4H10 = grams/molar mass
Using the coefficients in the balanced equation, convert mols C4H10 to mols H2O.
Now convert mols H2O to grams. g = mols H2O x molar mass H2O.