The balanced equation for the combustion of ethanol is 2C2H5OH (g) + 7O2 (g) → 4CO2 (g) + 6H2O (g) How many grams of oxygen gas are required to burn 5.54 g of C2H5OH? (write your answer with 3 sig figs and no units)

Question

Delacruz

Chemistry

11 April, 19:02

The balanced equation for the combustion of ethanol is 2C2H5OH (g) + 7O2 (g) → 4CO2 (g) + 6H2O (g) How many grams of oxygen gas are required to burn 5.54 g of C2H5OH? (write your answer with 3 sig figs and no units)

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Answers (1)

Know the Answer?

Fisher Blake · Answer 1

Answer: 13.5g of O2

Explanation:

2C2H5OH + 7O2 → 4CO2 + 6H2O

Molar Mass of C2H5OH = (12x2) + 5+16+1 = 46g/mol

Mass conc. Of C2H5OH = 2x46 = 92g

Molar Mass of O2 = 32g/mol

Mass conc. Of O2 = 7 x 32 = 224g

From the equation,

92g of C2H5OH required 224g O O2.

Therefore, 5.54g of C2H5OH will require = (5.54x224) / 92 = 13.5g of O2