Please somebody answer this question

Question

Given the following equation:

2 C₄H₁₀ + 13 O₂ →→ 8 CO₂ + 10 H₂O

How many grams of CO₂ are produced if 11.7 grams of C₄H₁₀ reacts with 54.6 grams of O₂? 

 g CO₂

What is the limiting reactant? 

1. there is no limiting reactant
2. O₂
3. C₄H₁₀
4. CO₂
5. H₂O

Answer 1

2C₄H₁₀ + 13O₂ ==> 8CO₂ + 10H₂O ... balanced equation for combustion of butane

Whenever you area given the amounts of BOTH reactants, you MUST first find the limiting reactant.

An easy way to do this is to divide the MOLES of each reactant by the corresponding coefficient in the balanced equation. Whichever value is lowest identifies the limiting reactant. For this problem, we have..

For C₄H₁₀: 11.7 g C₄H₁₀ x 1 mol C₄H₁₀ / 58.12 g = 0.201 mols C₄H₁₀ (÷2->0.10)

For O2: 54.6 g O₂ x 1 mol O₂ / 31.98 g = 1.71 mols O₂ (÷13->0.13)

The values of 0.10 and 0.13 are pretty close so they may consider that neither reactant is limiting, but if the math is correct, then since 0.10 is less than 0.13, C₄H₁₀ is the limiting reactant.

To find grams CO₂ produced, we now use the MOLES of the limiting reactant (0.201 mols C₄H₁₀) and the mole ratio in the balanced equation to find mols of CO₂, and then convert mols CO₂ to grams CO₂.

0.201 mols C₄H₁₀ x 8 mols CO₂ / 2 mols C₄H₁₀ = 0.804 mols CO₂

0.804 mols CO₂ x 44.0 g CO₂ / mol = 35.4 g CO₂ (3 sig. figs.)