C2H4 + 3O2 → 2CO2 + 2H2O At STP, 56.2L of C2H4 reacts with 42.6g of O2 What is the limiting reagent? What is the mass of CO2 expected?

Question

C2H4 + 3O2 → 2CO2 + 2H2O

At STP, 56.2L of C2H4 reacts with 42.6g of O2

1. What is the limiting reagent?
2. What is the mass of CO2 expected?

Answer 1

You are given 56.2 L of the ethane or C2H4. You can find out how many grams this is so you can make an equal comparison by using the density equation:

D= g/ml. Ethane density:. 0.3768

0.3758= g/56,200ml

If you solve you get 21,176.16g of ethane

Now that you have grams of each, do a mol given in problem to moles based on balanced equation,. The one that is bigger is not the limiting reactant the one that is smaller is.

.at weight of ethane: C2H4. 28.06g/mol

21,176.16g * 1mol/28.06g

=. 754.:67 mol of ethane

How many moles of O2 do we have?

42.6 g * 1mol/32g 1.33 mol of O2

1.33/3 = 0.4433. for O2.

Ethane. 754/1 = 754

Oxygen's number is a lot smaller, Oxygen is limiting this reaction based on the amount of ethane that was given in the problem. So now you can proceed the reaction with the Oxygen.

How much ethane results from the amount Oxygen?

1.33 mol O2 ==> 3 mol O2

X mol C2H4 ===> 1 mol C2H4

When you cross multiply and solve for x

3x = 1.33

X = 1.33/3. = 0.4433 moles of ethane

To find out how much CO2 follow the reaction with Oxygen.

3O2 === 2CO2

1.33 mol O2 ===> X CO2

2(1.33) = 3X

2(1.33)/3 = X =. 0.8867 mol of CO2

0.8867 mol * 44.01g/mol. = 39.02 g of CO2. 39.0 g CO2 for 3 sig figs.

This makes sense because when comparing the atomic weights of CO2 to O2. The grams are comparable and this is what would yield if given 42.6g of O2.

Answer 2

C₂H₄ + 3O₂ ==> 2CO₂ + 2H₂O ... balanced equation

An easy way to find the limiting reactant is to divide the moles of each reactant by the corresponding coefficient in the balanced equation and whichever value comes out less is the limiting reactant.

For C₂H₄: 56.2 L x 1 mol / 22.4 L @ STP = 2.51 mols (÷1-> 2.51)

For O₂: 42.6 g O₂ x 1 mol / 32 g = 1.33 mols (÷3->0.44)

O₂ is limiting since 0.44 is less than 2.51.

Now we use the moles of the limiting reactant to calculate moles and mass of CO2 expected.

1.33 mols O₂ x 2 mols CO₂ / 3 mols O₂ x 44 g CO₂/mol CO₂ = 39.0 g CO₂ expected