Question

the given reaction, if we react 7.4 grams of C4H10 and 27.8 grams of O2 , which is the limiting reactant?2 C4H10 + 13 O2 → 8 CO2 + 10 H2OSelect one:a.CO2b.H2Oc.C4H10d.O2

Answer 1

d. O₂

Step-by-step explanation

What is given:

Mass of C₄H₁₀ = 7.4 grams

Mass of O₂ = 27.8 grams

Equation: 2C₄H₁₀ + 13O₂ → 8CO₂ + 10H₂O

What to find:

To determine the limiting reactant.

Step-by-step solution:

Step 1: Determine the molar mass of C₄H₁₀ and O₂.

We shall use the atomic masses of H, C and O in the periodic table to calculate the molar mass of C₄H₁₀ and O₂.

Atomic masses of (H = 1.008, C = 12.011, O = 15.999)

Molar mass of C₄H₁₀ = (4 x 12.011) + (10 x 1.008) = 58.124 g/mol

Molar mass of O₂ = 2 x 15.999 = 31.998 g/mol

Step 2: Determine the moles of C₄H₁₀ and O₂ in 7.4 grams C₄H₁₀ and 27.8 grams O₂.

The moles of each reactant can be calculated using the mole formula below:

`Moles=\frac{Mass}{Molar\text{ }mass}`

So,

`\begin{gathered} Moles\text{ }of\text{ }C_4H_(10)=\frac{7.4\text{ }g}{58.124\text{ }g\text{/}mol}=0.13\text{ }mol \\ \\ Moles\text{ }of\text{ }O_2=\frac{27.8\text{ }g}{31.998\text{ }g\text{/}mol}=0.87\text{ }mol \\ \\ Moles\text{ }of\text{ }C_4H_(10)\text{ }to\text{ }O_2=(0.13mol)/(0.13mol):(0.87mol)/(0.13mol) \\ \\ Moles\text{ }ratio\text{ }of\text{ }C_4H_(10)\text{ }to\text{ }O_2=1:7 \end{gathered}`

Step 3: Determine the limiting reactant using the mole ratio from the given equation.

The mole ratio of C₄H₁₀ to O₂ from the given equation of reaction is 2 : 13.

This implies that 2 moles of C₄H₁₀ require 13 moles of O₂

From step 2, 1 mole of C₄H₁₀ reacted with 7 moles of O₂, therefore, O₂ is the limiting reactant.

The limiting reagent is the reactant that is completely used up in a reaction and thus determines when the reaction stops.

Thus, the correct answer is option d. O₂.