The balanced equation for the combustion of ethanol is
2C₂H₅OH(g) + 7O₂(g) → 4CO₂(g) + 6H₂O(g)
How many grams of dioxygen are required to burn 3.4 g of C₂H₅OH?

Question

Quizplus · Accepted Answer

The balanced equation shows that 2 moles of ethanol react with 7 moles of dioxygen. First, calculate the molar mass of ethanol (C2H5OH): 2(12.01) + 6(1.01) + 16.00 = 46.08 g/mol. Then, find how many moles 3.4 g of ethanol is: 3.4 g / 46.08 g/mol = 0.0738 mol. Since the reaction requires 7 moles of O2 for every 2 moles of ethanol, the moles of O2 needed are (7/2) * 0.0738 = 0.2589 mol. The molar mass of O2 is 32.00 g/mol, so the grams of O2 required are 0.2589 mol * 32.00 g/mol = 8.28 g, which rounds to 8.2 g.

The Balanced Equation for the Combustion of Ethanol Is