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!