If 34.1 grams of pentane (C5H12) are burned in excess oxygen, how many grams of H2O will be produced?
C5H12 + 8O2 ==> 5CO2 + 6H2O
mols pentane = grams/molar mass
Using the coefficients in the balanced equation, convert mols pentane to mols H2O
Now convert mols H2O to grams. g = mols x molar mass.
To determine the grams of H2O produced from the combustion of 34.1 grams of pentane (C5H12), we need to follow these steps:
Step 1: Write the balanced chemical equation for the combustion of pentane.
C5H12 + 8O2 → 5CO2 + 6H2O
Step 2: Determine the molar mass of pentane (C5H12).
Carbon (C) molar mass = 12.01 g/mol (5 atoms in pentane)
Hydrogen (H) molar mass = 1.01 g/mol (12 atoms in pentane)
Total molar mass of pentane = (5 × 12.01 g/mol) + (12 × 1.01 g/mol) = 72.15 g/mol
Step 3: Calculate the number of moles of pentane.
Number of moles = Mass of pentane / Molar mass of pentane
Number of moles = 34.1 g / 72.15 g/mol ≈ 0.472 mol
Step 4: Use stoichiometry to find the number of moles of water formed.
From the balanced equation, we see that for every 6 moles of H2O produced, we need 1 mole of pentane.
Number of moles of H2O = 0.472 mol × (6 mol H2O / 1 mol pentane) = 2.832 mol
Step 5: Calculate the mass of water produced.
Mass of H2O = Number of moles of H2O × Molar mass of H2O
Molar mass of H2O = (2 × 1.01 g/mol) + (1 × 16.00 g/mol) = 18.02 g/mol
Mass of H2O = 2.832 mol × 18.02 g/mol ≈ 51.0 g
Therefore, approximately 51.0 grams of H2O will be produced.
To determine the grams of H2O produced when burning pentane (C5H12), we need to balance the chemical equation and calculate the stoichiometry.
The balanced equation for the combustion of pentane is:
C5H12 + 8O2 → 5CO2 + 6H2O
From the balanced equation, we can see that for every 1 mole of pentane (C5H12) burned, 6 moles of water (H2O) are produced.
Step 1: Calculate the moles of pentane (C5H12) using the given mass and its molar mass.
Molar mass of pentane (C5H12):
(5 × atomic mass of carbon) + (12 × atomic mass of hydrogen)
Atomic mass of carbon = 12.01 g/mol
Atomic mass of hydrogen = 1.01 g/mol
Molar mass of pentane = (5 × 12.01 g/mol) + (12 × 1.01 g/mol)
Now, we can calculate the moles of pentane using its given mass:
Given mass of pentane = 34.1 g
Number of moles of pentane = Given mass / Molar mass
Step 2: Calculate the moles of water (H2O) produced using the stoichiometry.
According to the balanced equation, 1 mole of pentane (C5H12) produces 6 moles of water (H2O).
Number of moles of water (H2O) = Number of moles of pentane × 6
Step 3: Convert the moles of water (H2O) to grams using its molar mass.
Molar mass of water (H2O) = (2 × atomic mass of hydrogen) + (1 × atomic mass of oxygen)
Atomic mass of oxygen = 16.00 g/mol
Molar mass of water = (2 × 1.01 g/mol) + (1 × 16.00 g/mol)
Finally, we can calculate the grams of water (H2O):
Mass of water (H2O) = Number of moles of water × Molar mass of water
By following these steps, you can calculate the grams of water produced when 34.1 grams of pentane is burned in excess oxygen.