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Napoleon Bonaparte

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Using stoichiometry, we get: (Amount of grams of Fe₂O₃ decomposed) x (1 mol Fe₂O₃/ Molar mass of Fe₂O₃ in grams) x (824.2 kJ/2 mol Fe₂O₃) = 36.0 g Fe₂O₃ x (1 mol Fe₂O₃/ 159.69 g Fe₂O₃) x (824.2 kJ/2 mol Fe₂O₃) = (29.0 x
10.6 moles of CO2 are produced when 5.30 mol of ethane are burned in an excess of oxygen.
C is correct. We should choose the answer whose approximation is the closest to our question. Our question: 2.74/1.78 = 1.539325843 C = 8.70/5.65 = 1.539823009 The ratios between C and our question are very close.

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