Answer:
The balanced chemical equation for the combustion of ethane (C2H6) is:
C2H6 + O2 → CO2 + H2O
According to the stoichiometry of the balanced equation, 1 mole of C2H6 reacts with 3 moles of O2 to produce 2 moles of CO2. Therefore, the ratio of moles of O2 consumed to moles of CO2 produced is:
3 mol O2 / 2 mol CO2
To determine how many moles of O2 are consumed per every 50 moles of CO2 produced, we can set up a proportion:
3 mol O2 / 2 mol CO2 = x mol O2 / 50 mol CO2
Cross-multiplying and solving for x, we get:
x = (3 mol O2 / 2 mol CO2) x 50 mol CO2 = 75 mol O2
Therefore, 75 moles of O2 are consumed per every 50 moles of CO2 produced during the combustion of ethane.
Step-by-step explanation: