[Meeeeessstttehhhhh]

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!

[mjc123]

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.

[Meeeeessstttehhhhh]

THANK YOU!!!