Question

If 23.2 grams of butane (CaH10) and 93.7 grams of oxygen (O2) are available in the following reaction: C4H10 + O2 → CO2 + H2O a. Balance the equation for the reaction. 2 Cy Hiot 1302- 02 b. Determine which reactant is the limiting reactant. c. Calculate the theoretical yield of CO2 in grams. d. If the actual yield of CO2 is 69.2 g CO2, what is the percent yield?

Answer 1

(a.) The balanced reaction is : 2 C₄H₁₀ + 13 O₂ $\rightarrow$ 8 CO₂ + 10 H₂O

(b.) Case 1 : butane is the limiting reactant

mass C₄H₁₀ = 23.2 g

moles C₄H₁₀ = (mass C₄H₁₀) / (molar mass C₄H₁₀)

moles C₄H₁₀ = (23.2 g) / (58.12 g/mol)

moles C₄H₁₀ = 0.400 mol

moles CO₂ = (moles C₄H₁₀) * (8 moles CO₂ / 2 moles C₄H₁₀)

moles CO₂ = (0.400 mol) * (8 / 2)

moles CO₂ = (0.400 mol) * (4)

moles CO₂ = 1.60 mol

Case 2 : oxygen is the limiting reactant

mass O₂ = 93.7 g

moles O₂ = (mass O₂) / (molar mass O₂)

moles O₂ = (93.7 g) / (32.0 g/mol)

moles O₂ = 2.93 mol

moles CO₂ formed = (moles O₂) * (8 moles CO₂ / 13 moles O₂)

moles CO₂ formed = (2.93 mol) * (8/13)

moles CO₂ formed = 1.80 mol

Since less moles of CO₂ are formed in Case 1, therefore, butane is the limiting reactant.

(c.) Theoretical moles CO₂ formed = 1.60 mol

Theoretical yield CO₂ = (Theoretical moles CO₂ formed) * (molar mass CO₂)

Theoretical yield CO₂ = (1.60 mol) * (44.0 g/mol)

Theoretical yield CO₂ = 70.4 g

(d.) Percent yield = (actual yield / theoretical yield) * 100

Percent yield = (69.2 g / 70.4 g) * 100

Percent yield = (0.983) * 100

Percent yield = 98.3 %