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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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Eduard22sly

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

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