Question #110629

(a) A chemist only has 6. 50 grams of C2H2 and an unlimited supply of oxygen and he

desires to produce as much CO2as possible. If he uses the equation below, how much

oxygen should he add to the reaction?

2C2H2 (g ) + 5O2(g) → 4CO2(g) + 2 H2O(l)

Expert's answer

  1. Covert grams of C2H2 in moles

6.50gC2H2(1moleC2H226.04gC2H2)=0.250molC2H26.50 g C_2H_2 (\frac{1 mole C_2H_2}{26.04 g C_2H_2})=0.250 mol C_2H_2


2.Convert moles of C2H2 to moles of O2:

0.250molesC2H2(5molesO22molesC2H2)=0.625molesO20.250 moles C_2H_2 (\frac{5 moles O_2}{2 moles C_2H_2}) = 0.625 moles O_2


3.Convert moles of O2 to mass of O2

0.625molesO2(32.00gO21moleO2)=20.0gO20.625 moles O_2(\frac{32.00 g O_2}{1 mole O_2})=20.0 g O_2


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