Ethyl acetate reacts with H2 in the presense of a catalyst to yield ethyl alcoho

Question

Ethyl acetate reacts with H2 in the presense of a catalyst to yield ethyl alcohol: C4 H8O2(l) +H2(g)-------> C2H6O6(l)

(a) Write a balanced equation for the reaction

(b) How many moles of ethyl alcohol are produced by reaction of 1.5 mol of ethyl acetate?

(c) How many grams of ethyl alcohol are produced by reaction 1.5 mol of ethyl acetate with H2?

(d) How many grams of ethyl alcohol are of ethyl alcohol are produced by reaction of 12.0 g of ethyl acetate with H2?

(e) How many grams of H2 are needed to react with 12.0g of ethyl acetate?

Explanation / Answer

C4H8O2 +2 H2 = 2 C2H6O

moles C2H6O= 1.5 x2=3.0

3.0 mol x 46 g/mol=138 g

moles C4H8O2 = 12.0/88 g/mol=0.1363
moles C2H6O = 0.1363x2=0.27272
mass = 0.27272 x 46=12.5454 g

moles H2 needed =0.1363x2= 0.2726
mass H2 = 0.2726 mol x 2 g/mol=0.5452 g

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