Answer:
a) 8.4 moles ethyl alcohol
b) 387 grams ethyl alcohol
c) 16.77 grams ethyl alcohol
d) 0.735 grams H2
Step-by-step explanation:
Step 1: Data given
Molar mass of ethyl acetate = 88.11 g/mol (C4H8O2)
Molar mass of ethyl alcohol = 46.07 g/mol (C2H6O)
Step 2: The balanced equation
C4H8O2 + 2H2 → 2C2H6O
Step 3: How many moles of ethyl alcohol are produced by reaction of 4.2 mol of ethyl acetate?
For 1 mol mol ethyl acetate we need 2 moles H2 to produce 2 moles ethyl alcohol.
For 4.2 mol of ethyl acetate we'll have 2*4.2 = 8.4 moles ethyl alcohol
Step 4: How many grams of ethyl alcohol are produced by reaction of 4.2 mol of ethyl acetate with H2 ?
Mass of ethyl alcohol = moles * molar mass
Mass ethyl alcohol = 8.4 moles * 46.07 g/mol
Mass ethyl alcohol = 387 grams
Step 5: How many grams of ethyl alcohol are produced by reaction of 16.0 g of ethyl acetate with H2 ?
Moles ethyl acetate = 16.0 grams / 88.11 g/mol = 0.182 moles
Moles ethyl alcohol = 2*0.182 = 0.364 moles
Mass ethyl alcohol = 16.77 grams
Step 6: How many grams of H2 are needed to react with 16.0 g of ethyl acetate?
For 0.182 moles ethyl acetate we need 2*0.182 = 0.364 moles H2
0.364 moles H2 = 0.364 * 2/02 g/mol = 0.735 grams H2