Item 9 Part A Determine the limiting reactant for the reaction. (Hint: Write a b

Question

Item 9 Part A Determine the limiting reactant for the reaction. (Hint: Write a balanced equation for the combustion of ethanol.) The combustion of liquid ethanol (C2Hs OH) produces carbon dioxide and water. After 4.64 mL of ethanol (density = 0.789 g/ml) was allowed to bum in the presence of 15.35 g of oxygen gas, 3.72 mL of water (density- 1.00 g/ml) was collected. ethanol oxygen Submit My Answers Give Up Correct Part B Determine the theoretical yield of H2O for the reaction. rra Submit My Answers Give Up Incorrect; Try Again; 4 attempts remaining Part C Determine the percent yield of H2 O for the reaction.

Explanation / Answer

A) the reaction is

C2H5OH + 302 ---> 2 C02 + 3 H20

we know that

mass = density x volume

so

mass of C2H5OH = 0.789 x 4.64 = 3.66096

now

moles = mass / molar mass

so

moles of C2H5OH = 3.66096 / 46

moles of C2H5OH = 0.079586

now

moles of oxygen = 15.35 / 32

moles of oxygen = 0.48

now

moles of ethanol reuqired = ( 1/3) x moles of oxygen

moles of ethanol required = 0.48 / 3 = 0.16

but only

0.079 moles of ethanol is present

so

ethanol is the limitng reagent

B)

now

C2H5OH + 302 ---> 2 C02 + 3 H20

we can see that

theoretical moles of H20 = 3 x moles of C2H5OH

so

theoretical moles of H20 = 3 x 0.079586

theoretical moles of H20 = 0.238758

now

theoretical masss of H20 = 0.238758 x 18

theoretical mass of H20 = 4.2976

so

theoretical yield of H20 is 4.2976 grams

C)

given

actual yield of H20 = 3.72 grams

now

% yield = actual x 100 / theoretical

% yield = 3.72 x 100 / 4.2976

% yield = 86.56

so

the percent yield of H20 is 86.56 %

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