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?

(a) You may find the balanced chemical equation in the attached picture, where 1 mole of ethyl acetate will react with 2 moles of hydrogen to produce 2 moles of ethyl alcohol.

(b) Knowing the balanced chemical equation, we devise the following reasoning:

if 1 mole of ethyl acetate produces 2 moles of ethyl alcohol

then 1.5 moles of ethyl acetate produces X moles of ethyl alcohol

X = (1.5 × 2) / 1 = 3 moles of ethyl alcohol

(c) From the point (b) we know that 1.5 mol of ethyl acetate produces 3 moles of ethyl alcohol.

mass = number of moles × molecular weight

mass of ethyl alcohol = 3 × 46 = 138 g

(d) number of moles = mass / molecular weight

number of moles of ethyl acetate = 12 / 88 = 0.136 moles

Now from the balanced chemical reaction we deduce that:

if 1 mole of ethyl acetate produces 2 moles of ethyl alcohol

then 0.136 moles of ethyl acetate produces X moles of ethyl alcohol

X = (0.136 × 2) / 1 = 0.272 moles of ethyl alcohol

mass = number of moles × molecular weight

mass of ethyl alcohol = 0.272 × 46 = 12.5 g

(e) We know from point (d) that 12 g of ethyl acetate are equal to 0.136 moles of ethyl acetate.

Taking in account the chemical reaction, we know that:

if 1 mole of ethyl acetate react with 2 moles of hydrogen

then 0.136 moles of ethyl acetate will react with Y moles of hydrogen

Y = (0.136 × 2) / 1 = 0.272 moles of hydrogen

mass = number of moles × molecular weight

mass of hydrogen (H₂) = 0.272 × 2 = 0.544 g

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