1. Acetic acid contains the elements carbon, hydrogen, and oxygen. When 5.00 g o

Question

1. Acetic acid contains the elements carbon, hydrogen, and oxygen. When 5.00 g of acetic acid are burned in air, 7.33g of CO2, and 3.00g of water are obtained. Determine the empirical formula of acetic acid. (5 points)

2.

For the equation,                Al    +      O2           ----->                         Al2O3             answer the following questions: (5 points)

a. Balance the above equation

b. If 10.0 g of Al are used, how many g of oxygen are needed and how many g of product are produced?

c. If you had 20 moles of Al, how many moles of oxygen would be needed and how many moles of product would be produced?

d. Based on the information in b., if you isolated 5.0 g of Al2O3, calculate the % yield.

                 e. If you had 5.0 g of Al and 4.5 g of O2, which is the limiting reagent?

Explanation / Answer

Q1 Solution

Given data

5.00 g acetic acid

7.33 g CO2

3.00 g H2O

Empirical formula = ?

Lets first calculate the moles of the CO2 and H2O

Moles = mass / molar mass

Moles of CO2 = 7.33 g / 44.01 g per mol   = 0.1666 mol CO2

Since mole ratio of the CO2 to C is 1 :1 so moles of C = 0.1666 mol

Moles of H2O = 3.00 g / 18.0148 g per mol = 0.1666 mol H2O

Since mole ratio of the H2O to H is 1 :2 therefore moles of H = 0.1666 * 2 = 0.3332 mol H

Now lets calculate mass of C and H

Mass = moles * molar mass

Mass of C = 0.1666 mol * 12.01 g per mol = 2.00 g

Mass of H =0.3332 mol * 1.0079 g per mol = 0.336 g H

Now lets calculate mass of O in the sample

Mass of oxygen = total mass – (mass of C+ mass of hydrogen )

Mass of oxygen = 5.00 g – (2.00 g + 0.336 g) = 2.664 g O

Lets calculate moles of Oxygen

Moles of O = 2.664 g / 16.0 g per mol = 0.1665 mol O

Now lets find the ratio of the moles

Diovide each mole value by the smallest mole value

C= 0.1666 / 0.1665 = 1

H= 0.3336 / 0.1665 = 2

O = 0.1665 / 0.1665 = 1

Therefore the empirical formula of the acetic acid = CH2O

Q2 ) a) balanced reaction equation is as follows

4Al +3O2 -----> 2Al2O3

b) to calculate the mass of oxygen needed to react with 10.0 g Al and to calculate the amount of the product forming we need to use the mole ratio from the balanced reaction equation.

Lets calculate mass of O2 needed

Mole ratio of the Al to O2 is 4 :3

(10.0 g Al * 1 mol / 26.98 g ) * (3 mol O2/4 mol Al) *(32.0 g / 1 mol O2) = 8.89 g O2

Lets calculate mass of the Al2O3 produced

(10.0 g Al * 1 mol / 26.98 g )*(2 mol Al2O3 / 4 mol Al ) * (101.96 g/ 1 mol Al2O3) = 18.9 g Al2O3

c) 20 mol Al

mole of O2 = ?

mole of Al2O3 = ?

using the mole ratio of O2 and Al2O3 with the Al we have to calculate moles of the O2 and Al2O3

lets calculate moles of O2            

20 mol Al * 3 mol O2 / 4 mol Al = 15 mol O2

Lets calculate moles of Al2O3

20 mol Al * 2 mol Al2O3 / 4 mol Al = 10 mol Al2O3

d) actual yield = 5.00 g

theoretical yield from part b = 18.9 g

% yield = (actual value / theoretical value ) * 100 %

             = (5.00 g / 18.9 g)*100%

             = 26.46 %

e) given 5.0 g Al and 4.5 g O2

limiting reactant = ?

moles of Al = 5.0 g / 26.98 g per mol = 0.1853 mol Al

moles of O2 = 4.5 g / 32 g per mol = 0.141 mol O2

lets calculate moles of O2 needed to react with 0.1853 mol Al using their mole ratios

0.1853 mol Al * 3 mol O2/ 4 mol Al = 0.139 mol O2

Since moles of O2 are more than actually needed to react with 0.1853 mol Al

Therefore Al is the limiting reactant

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