Determine the number of grams of C4H10 that are required to completely react to produce 8.70 mol of CO2 according to the following combustion reaction: 2C₄H₁₀(g) + 13O₂(g) →8 CO₂(g) +10 H₂O(g)

Question

I can't do this problem and I've been trying to figure out the conversion factors for ages but I'm stuck

Answer 1

Good news is that you realize that this is a conversion problem where you are trying to get from point A, 8.70 moles of CO2, to point B, number of grams of C₄H₁₀. However, these are always difficult for students but after much practice, you will finally understand them and ace that test.

For now, let's work through this problem. When working problems such as this, I was taught that you try to cancel out the units.

8.70 moles CO₂ This is where we start. We look at the equation below and notice that for 8 mole of CO₂, there is 2 moles of C₄H₁₀. This will be our first conversion factor which will allow us to cancel out moles of CO₂.

8.70 moles CO₂ x (2 moles of C₄H₁₀/8 moles of CO₂) = 2.175 moles of C₄H₁₀ This gives us moles of C₄H₁₀. Now to calculate grams of C₄H₁₀, we need the molecular weight of this.

Carbon - 12.01 There are four Carbons so 4(12.01) = 48.04 grams.

Hydrogen - 1.008 There are 10 Hydrogens so 10(1.008) = 10.08 grams.

So the molecular weight of C4H10 is 10.08 grams + 48.04 grams = 58.12 grams.

This tells us that for every mole of C₄H₁₀, there is 58.12 grams. This will be our last conversion factor.

2.175 moles of C₄H₁₀ x (58.12 grams/1 mole) = 126.411 grams.

Using appropriate significant figures, this gives us 126 grams.