Stoichiometry!
Please note:
- Use 6.022x1023 for avogadro’s number
- Ignore sig figs and do not round the final answer.
- Keep it to 1 decimal place.

Hey there!

C₃H₆O(l) + O₂(g) => CO₂(g) + H₂O(g)

First thing I want to do is balance the equation. Balance C:

C₃H₆O(l) + O₂(g) => 3CO₂(g) + H₂O(g)

Balance H:

C₃H₆O(l) + O₂(g) => 3CO₂(g) + 3H₂O(g)

Balance O:

C₃H₆O(l) + 4O₂(g) => 3CO₂(g) + 3H₂O(g)

Now we can properly solve the problem.

a.)

Density of acetone is 0.800 g/mL, we need to form 67.2 L of CO₂ at STP.

For every one mole of acetone reacted, 3 moles of CO₂ is produced.

Let's convert 67.2 L to moles: At STP, one mole of a gas takes up 22.4 liters.

67.2 ÷ 22.4 = 3 moles

So turns out we want to produce 3 moles of CO₂, which means we need one mole of acetone.

The molar mass of acetone is 58.08 g/mol. The density of acetone is 0.800 g/mL. We need the volume of one mole.

58.08 ÷ 0.800 = 72.6

72.6 mL of acetone is needed.

b.)

For every one molecule of acetone reacted, 3 molecules of water is produced.

We are combusting 3.011 x 10²² molecules of acetone.

3.011 x 10²² x 3 = 9.033 x 10²²

Find the number of moles:

(9.033 x 10²²) ÷ (6.022 x 10²³) = 0.15 moles

The molar mass of water is 18.015 g/mol.

0.15 x 18.015 = 2.7 grams

2.7 grams of water vapor is formed.

Hope this helps!