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200 grams of C3H4 gas with 20% purity is burned with excess O2 gas. How many grams of H2O are
produced with 60% yield?


Sagot :

Answer:

Explanation:

approx 30%

Reaction

C₃H₄ + 4O₂ → 3CO₂ + 2H₂O

20% purity = 20% x 200 = 40 g

mole C₃H₄ = mass : molar mass = 40 : 40 g/mole = 1

mole H₂O from reaction coefficient (C₃H₄ limiting reactant) = 2/1 x 1 mole = 2 mole

mass H₂O = mole x molar mass = 2 x 18 g/mole = 36 g

for 60% yield, H₂O produced  = 60% x 36 g = 21.6 g