Limiting Reactants: Compare Product Method (Mass/Mass)

Question

For the following reaction, 8.90 grams of butane (C₄H₁₀) are allowed to react with 19.0 grams of oxygen gas.

butane (C₄H₁₀) (g) + oxygen (g) carbon dioxide (g) + water (g)

What is the maximum amount of carbon dioxide that can be formed?  ____grams

What is the FORMULA for the limiting reagent?

What amount of the excess reagent remains after the reaction is complete?  ____grams

Answer 1

Omar, maybe this will help clarify things.

The easiest way to find the limiting reactant is to simply divide the moles of each reactant by the corresponding coefficient in the balanced equation. So it is imperative to have a correctly balanced equation:

2C₄H₁₀ + 13O₂ ==> 8CO₂ + 10H₂O

Now to find limiting reactant:

moles C₄H₁₀ present = 8.9 g x 1 mol C4H10 / 58 g = 0.153 moles C₄H₁₀ (÷2->0.077)

moles O₂ present = 19.0 g O₂ x 1 mol O₂ / 32 g = 0.594 moles O₂ (÷13=>0.0457) THIS IS LIMITING (O₂)

The limiting reactant will dictate how much product can be formed:

0.594 moles O₂ x 8 mol CO₂ / 13 mol O₂ x 44 g CO₂/mole CO₂ = 16.1 g CO₂ that can be formed

To find how much of the excess reagent (C₄H₁₀) remains, we simply find out how much was used, and then subtract that from how much we started with (0.153 moles)

How much was used?

0.594 mol O₂ x 2 mol C₄H₁₀ / 13 moles O₂ = 0.0914 moles C₄H₁₀ used

moles C₄H₁₀ remaining = 0.153 moles - 0.0914 moles = 0.0616 moles C₄H₁₀ remaining

mass C₄H₁₀ remaining = 0.0616 moles x 58 g/mol = 3.57 g C₄H₁₀ remains

Answer 2

The reaction is 2C₄H₁₀+13O₂=8CO₂+10H₂O

molar mass of C₄H₁₀ is 58.12

molar mass of O₂ is 32

The reactants' mass ratio C₄H₁₀/O₂=2*58.12/(13*32)=0.279

8.90g C₄H₁₀ needs O₂ 8.90/0.279=32g>19g

so C₄H₁₀ is excessive

13*32g O₂ generates 8*44g CO₂, so 19g O₂ generates CO₂ 19*8*44/(13*32)=16.08g

19g O₂ needs C₄H₁₀ 19*2*58.12/(13*32)=5.31g to finish the reaction

8.90-5.31=3.59g C₄H₁₀ remains after finishing the reaction