Question

5. Consider the combustion of carbon monoxide (CO) in oxygen gas: 2CO(g) + O₂(g) → 2CO₂(g)

A. Calculate the number of moles of CO₂ produced if 2.86 moles of CO is reacted with excess oxygen._______ moles CO_2.

B. Determine the moles of oxygen required to react with the CO._______ moles O₂

6. Consider the following balanced chemical equation: C₂H₅OH(l) + 3O₂(g) → 2CO₂(g) + 3H₂O(l)

a. How many moles of CO₂ form when 1.73 moles of C₂H₅OH react? ______ moles CO₂

b. How many moles of H₂O form from the reaction of 1.73 moles of C₂H₅OH?_____ moles H₂O

c. How many moles of CO₂ form when 1.73 moles of H₂O form? _____moles CO₂

d. How many molecules of CO₂ form when 1.73 moles of H₂O form? _____ × 10 ^? molecules CO₂

7.   How much oxygen (in grams) will form from 15.2 mg of N₂O₅?

2N₂O₅(s) → 4NO₂(g) + O₂(g)---------> × 10^?g O₂

8. How much oxygen (in grams) will form from 65.2 g of N₂O₅?

2N₂O₅(s) → 4NO₂(g) + O₂(g) ---------> × 10^? g O₂

Answer 1

Part 5

A.

We are given the reaction

. It is clear from the reaction that 2 mols of CO will react with 1 mol of Oxygen to give 2 mols of Carbon dioxide.

Now if 2.86 mols of CO are given, we are supposed to find out the number of mols of CO₂ produced.

It is very clear from the reaction that 2mols of CO produce 2 mols of CO₂. This implies that there is a 1:1 relationship between them. ie, the number of mols of CO equal the number of moles of CO₂ produced.

Hence the number of mols of CO₂ produced is 2.86mols

The key to answering such question is a careful study of the reaction equation. It is very important to pay attention to the stoichiometric coefficients.

B.

In part A, we had excess O₂ hence did not have to worry about the number of moles of it required for the reaction. But it can clearly be seen that 2 mols of CO react with 1mol of Oxygen. There exists a 2:1 ratio between them.

Hence, by using simple maths, the number of mols of Oxygen required would be 2.86/2 = 1.43mols.

Question 6

Observe the stoichiometric ratios very carefully. It will help in answering all the subparts very very easily.

A.

The ratio between C₂H₅OH and CO₂ is 1:2, ie one mole of ethanol gives 2 mol of CO₂ . Hence it is trivial that 1.73 mols of C₂H₅OH will give 1.73*2 = 3.46mols of CO_2.

B.

The stoichiometric ratio is 1:3 in this case. using the same approach as the part A, we say that 1.73mols of C₂H₅OH will give 1.73*3 = 5.19 mols of water.

C.

Now this part is tricky. But can be answered very easily using the stoichiometric ratio again. We just need to ratio between CO₂ and H₂O. This ratio is 2:3.

ie if 2 mols of CO₂ are formed then 3 mols of H₂O are formed.

If 1.73 mols of H₂O are formed, then, the number of CO₂ is = (2*1.73)/3 = 1.153 mols.

Part C.

The number of molecules of CO₂ can be got by multiplying the number of mols of CO₂ by the Avogadros constant.

We get

Number of molecules = 6.023*10²³*1.153 = 6.946*10²³ Molecules of CO₂

Question 7

Given reaction is

.

The stoichiometric ratio is . 2 mols react to give 1 mol of Oxygen.

15.2mg of are nothing but . Hence the number of mols of Oxygen formed are

.

(We have use the formulae for number of mols = wt/Molecular weight)

Question 8.

This is the same as Question 7

Number of mols of N₂O₅ is .

Hence the number of mols of Oxygen formed are

I hope this has helped. In case the asnwer is unclear, feel free to leave a comment. I will be glad to help you out.