Question

if 7.51 moles co₂ are emitted, how many grams of octane fuel (c₈h₁₈) does this engine have to burn to meet the requirement?
2c₈h₁₈ (l) + 25o₂ (g) → 16co₂ (g) + 18h₂o (l)
pick one setup and drag you the appropriate values into those boxes.
dimensional analysis:
boxes
ratios:
boxes = boxes
options: 7.51 mol co₂, x mol c₈h₁₈, 1 mol co₂, 8 mol co₂, 16 mol co₂, 1 mol c₈h₁₈, 2 mol c₈h₁₈, 6.02 x 10²³ molecules co₂, 6.02 x 10²³ molecules c₈h₁₈, 44 g co₂, 114 g c₈h₁₈, 1, 2, 8, 16, 0.00823 g c₈h₁₈, 107 g c₈h₁₈

Explanation:

Step1: Molar ratio of $\text{C}_8\text{H}_{18}$ to $\text{CO}_2$

From the balanced equation: $\frac{2\ \text{mol}\ \text{C}_8\text{H}_{18}}{16\ \text{mol}\ \text{CO}_2}$

Step2: Calculate moles of $\text{C}_8\text{H}_{18}$

$7.51\ \text{mol}\ \text{CO}_2 \times \frac{2\ \text{mol}\ \text{C}_8\text{H}_{18}}{16\ \text{mol}\ \text{CO}_2} = 0.93875\ \text{mol}\ \text{C}_8\text{H}_{18}$

Step3: Molar mass of $\text{C}_8\text{H}_{18}$

Molar mass = $8\times12.01 + 18\times1.008 = 114.224\ \text{g/mol}$

Step4: Calculate mass of $\text{C}_8\text{H}_{18}$

$0.93875\ \text{mol}\ \text{C}_8\text{H}_{18} \times 114.224\ \text{g/mol} \approx 107\ \text{g}$

Dimensional Analysis Setup:

Top row (left to right): $7.51\ \text{mol}\ \text{CO}_2$, $2\ \text{mol}\ \text{C}_8\text{H}_{18}$, $114\ \text{g}\ \text{C}_8\text{H}_{18}$
Bottom row (left to right): $16\ \text{mol}\ \text{CO}_2$, $1\ \text{mol}\ \text{C}_8\text{H}_{18}$

Ratios Setup:

$\frac{2\ \text{mol}\ \text{C}_8\text{H}_{18}}{16\ \text{mol}\ \text{CO}_2} = \frac{x\ \text{mol}\ \text{C}_8\text{H}_{18}}{7.51\ \text{mol}\ \text{CO}_2}$
Then use molar mass: $\text{Mass} = x\ \text{mol} \times 114\ \text{g/mol}$

Answer:

107 grams of octane fuel ($\text{C}_8\text{H}_{18}$) need to be burned.