If 20.0g acetylene is allowed to completeyl react with o@ how many g of O2 are reacted? 2C2H2 +5O2----4CO2 +2H20
1. Figure the moles of acetylene
Now look at the balanced equation. YOu need 5/2 moles of O2 for every one mole of acy. So you must need 5/2 * moles of acetylene. Now convert that to grams.
To find out how many grams of O2 are reacted when 20.0g of acetylene (C2H2) is completely reacted, we need to use the stoichiometry of the balanced chemical equation.
The balanced equation is:
2C2H2 + 5O2 → 4CO2 + 2H2O
The coefficient in front of O2 is 5, which means that for every 5 moles of O2, 2 moles of acetylene are reacted.
First, we need to determine the number of moles of acetylene (C2H2) in 20.0g. To do this, we use the molar mass of acetylene, which is 26.04 g/mol.
Number of moles of C2H2 = mass / molar mass = 20.0g / 26.04 g/mol ≈ 0.768 mol
According to the balanced equation, 2 moles of acetylene react with 5 moles of O2.
So, the number of moles of O2 needed is:
Number of moles of O2 = (5/2) × number of moles of C2H2 = (5/2) × 0.768 mol ≈ 1.92 mol
Finally, we can determine the mass of O2 using its molar mass, which is approximately 32.00 g/mol.
Mass of O2 = number of moles × molar mass = 1.92 mol × 32.00 g/mol ≈ 61.44 g
Therefore, when 20.0g of acetylene completely reacts with O2, approximately 61.44 grams of O2 are reacted.