Answer to Question #111233 in General Chemistry for Sinovuyo

Question

The yield of the reaction is 91.5% and the purity of the oxygen gas is 78.5%.
250.0g of impure oxygen gas react with an excess amount of octane.
Calculate the mass (in grams) of water vapour that will be formed.
Equation: 2C8H18 + 25O2-> 16CO2 + 18H2O(g)

Answer 1

Find the mass of pure oxygen by the formula:

m₂(O₂)= (m₁(O₂)*$\omega$(O₂))/100%

m₁-impure oxigen

$\omega$ - purity of the oxygen

m₂(O₂)=(250g*78,5%)/100%=195,25g

Find the amount of substance oxygen:

n(O₂)=m₂(O₂)/M(O₂) = 196,25g/32g/mol=6,13mol

Then, according to the reaction equation, the amount of substance water will be equal to:

n(H₂O)=(6,13mol*18mol)/25mol=4,41mol

Then, theoretical mass of water:

m₁(H₂O)=n(H2O)*M(H2O)=4,41mol*18g/mol=79,38g

The practical mass of water:

m₂(H₂O)=(79,38g*91,5%)/100% = 72,63g