Which reagent will be used up first if 78.1 g of O₂ is reacted with 62.4 g of C₄H₁₀? 2 C₄H₁₀ + 13 O₂ → 8 CO₂ + 10 H₂O

Question

Quizplus · Accepted Answer

To solve the problem, we need to determine which of the reactants will limit the amount of product that can be formed. This is called the limiting reagent. To find the limiting reagent, we need to use stoichiometry to calculate how much product can be formed from each reactant.

Let's start by finding the moles of each reactant:

moles of O₂ = 78.1 g / 32 g/mol = 2.44 mol
moles of C₄H₁₀ = 62.4 g / 58 g/mol = 1.08 mol

Next, we need to use the coefficients in the balanced chemical equation to determine how many moles of product can be formed from each reactant:

From 2 moles of C₄H₁₀, we can form 8 moles of CO₂ 
From 13 moles of O₂, we can form 10 moles of H₂O

Therefore, we can see that we will form more CO₂ from the butane than from the oxygen.

Using the moles of each reactant, we can calculate how much product can be formed:

CO₂ from butane = 8 mol CO₂/ 2 mol C₄H₁₀ * 1.08 mol C₄H₁₀ = 4.32 mol CO₂
CO₂ from oxygen = 8 mol CO₂/ 13 mol O₂ * 2.44 mol O₂ = 1.49 mol CO₂

H₂O from butane = 10 mol H₂O/ 2 mol C₄H₁₀ * 1.08 mol C₄H₁₀ = 5.4 mol H₂O
H₂O from oxygen = 10 mol H₂O/ 13 mol O₂ * 2.44 mol O₂ = 1.88 mol H₂O

We can see that we will form less CO₂ and less H₂O from the oxygen than from the butane, which means that the oxygen is the limiting reagent. Therefore, the answer is B) oxygen.

Which Reagent Will Be Used Up First If 78