2 C 4 H 10 + 13 O 2 → 8 CO 2 + 10 H 2 O -Which reagent will be used up first if 78.1 g of O₂ is reacted with 62.4 g of C₄H₁₀?

Question

Quizplus · Accepted Answer

To determine which reagent will be used up first, we need to use stoichiometry and compare the moles of each reagent to the amount of product that can be formed. First, we need to balance the equation to determine the molar ratios: 2 C₄H₁₀ + 13 O₂ → 8 CO₂ + 10 H₂O Next, we will convert the masses of each reagent to moles: 62.4 g C₄H₁₀ x 1 mol / 214.34 g = 0.291 mol C₄H₁₀ 78.1 g O₂ x 1 mol / 268.38 g = 0.291 mol O₂ Whichever reagent has the smallest number of moles is the limiting reagent and will be used up first. In this case, both reagents have the same number of moles, so we need to compare the amount of product that can be formed from each reagent. Using the balanced equation, we can see that 13 moles of oxygen can produce 8 moles of CO₂, while only 2 moles of C₄H₁₀ can produce 8 moles of CO₂. This means that the amount of product that can be formed from the oxygen is much greater than the amount that can be formed from the C₄H₁₀. Therefore, oxygen is not the limiting reagent and will not be used up first. The correct answer is B) oxygen.

2 C4H10 + 13 O2 → 8 CO2 + 10

2 C₄H₁₀ + 13 O₂ → 8 CO₂ + 10