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Author Topic: WANTED: Approval of stoichiometry homework  (Read 844 times)

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Meeeeessstttehhhhh

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WANTED: Approval of stoichiometry homework
« on: January 03, 2018, 06:51:10 AM »

Hey everyone!

I would really appreciate it if someone could give a quick look over my homework, and confirm that it is in fact correct. If you do I will be eternally grateful!

Question 1: When propane (C3H8) is burned in oxygen, the products are carbon dioxide and water. Write a balanced equation for this reaction and identify the mole ratios between propane and oxygen, and between propane and carbon dioxide.

Answer: C3H8 + 5O2 -> 3CO2 + 4H2O
1 mol C3H8 reacts with 5 mol O2       C3H8:O2 = 1:5
1 mol C3H8 produces 3 mol CO2         C3H8:CO2 = 1:3

Question 2: What mass of carbon dioxide is produced by the combustion of 50.0 g of propane?

Answer:
Balanced equation:
   C3H8 + 5O2 -> 3CO2 + 4H2O
Number of moles in C3H8:
   =(50.0 g)/((3 ×12.01 g/mol) + (8 ×1.01 g/mol))
        =1.133529812 mol
Mole ratio of C3H8 to CO2:
   1:3
There are 1.133529812 mol x 3 = 3.400589436 mol of CO2 produced.
Mass of CO2 = 3.400589436 mol x 44.01 g/mol
                   = 149.6599411 g CO2
Therefore, the carbon dioxide produced by the combustion of 50.0 g of propane has a mass of 150 g.

Question 3: Aluminum metal is produced from the electrical decomposition of aluminum oxide found in a natural ore called bauxite. How much aluminum oxide is needed to produce 1.000 kg of aluminum?

Answer:
Balanced equation:
   2Al2O3 -> 4Al + 6O2
Number of moles in Al:
   =(1000 g)/(26.98 g/mol)
   =37.06449222  mol
Mole ration of Al to Al2O3:
   4:2 or 2:1
There are 37.06449222 mol x (1/2) = 18.53224611 mol of Al2O3 required.
Mass of Al2O3 = 18.53224611 mol x 101.96 g/mol
                     = 1889.547813 g
Therefore, 1.890 kg of aluminum oxide is needed to produce 1.000 kg of aluminum.

Question 4: Calculate what mass of oxygen is required to completely combust 500.0 g of gasoline. Assume that the gasoline contains only octane.

Answer:
Balanced equation:
   2C8H18 + 25O2 -> 16CO2 + 18H2O
Number of moles in C8H18:
   =(500.0 g)/((8 × 12.01 g/mol) + (18 × 1.01 g/mol) )
   =4.375984597 mol
Mole ratio of C8H18 to O2:
   2:25
There are 4.375984597 mol x (2/25) = 54.69980746 mol of O2 required.
Mass of O2 = 54.69980745 mol x 32 g/mol
                 =1750.393839 g O2
Therefore, 1.750 kg of O2 are required to fully combust 500.0 g of gasoline.

Question 5: How many grams of vinegar are required to react completely with 25.0 g of baking soda?

Answer:
Balanced equation:
   HC2H3O2 + NaHCO3 -> NaC2H3O2 + H2O + CO2
Number of moles in NaHCO3:
    =(25.0 g)/(22.99 g∕mol+ 1.01 g∕mol+ 12.01 g∕mol+(3 × 16.00 g∕mol))
    =0.297583621 mol 
Mole ratio of NaHCO3 to HC2H3O2:
   1:1
There are 0.297583621 mol x 1 = 0.297583621 mol of HC2H3O2 required.
Mass of HC2H3O2 = 0.297583621 mol x 60.06 g/mol
                           = 17.87287228 g HC2H3O2
Therefore, 17.9 g of vinegar are required to react completely with 25.0 g of baking soda.

Thank you so very much in advance!
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mjc123

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Re: WANTED: Approval of stoichiometry homework
« Reply #1 on: January 03, 2018, 07:09:56 AM »

I haven't checked all the arithmetic, but your approach looks right.
Question 2: if you spot that the molecular weights of propane and CO2 are both 44, then the mass ratio equals the mole ratio. But your procedure is right in the general case.
Question 3: balanced equation requires 3O2, not 6, but it doesn't affect the answer.
Question 4: moles octane should be multiplied by 25/2, not 2/25, but you got the right answer.
Question 5: "vinegar" is of course not pure acetic acid, but in the absence of any information on the strength of the "vinegar", you can only do what you have done.
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Meeeeessstttehhhhh

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Re: WANTED: Approval of stoichiometry homework
« Reply #2 on: January 03, 2018, 07:22:26 AM »

THANK YOU!!!
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