Answer:
73.48 g
Step-by-step explanation:
The reaction that takes place is:
- 2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O
We convert the given reactant masses to moles, using their respective molar masses:
- 25 g ethane ÷ 30 g/mol = 0.83 mol ethane
- 122 g O₂ ÷ 32 g/mol = 3.81 mol O₂
0.83 moles of ethane would react completely with (0.83*7/2) 2.9 moles of O₂. There are more O₂ moles than that, so O₂ is the reactant in excess and ethane is the limiting reactant.
We calculate the produced moles of CO₂ using the moles of the limiting reactant:
- 0.83 mol C₂H₆ *
= 1.67 mol CO₂
Finally we convert CO₂ moles to grams, using its molar mass:
- 1.67 mol CO₂ * 44 g/mol = 73.48 g