Question

answer the questions below given the balanced chemical equation and the following quantities: 153.8 grams c₃h₈ 352 grams o₂ 1 c₃h₈ + 5 o₂ → 3 co₂ + 4 h₂o the balanced chemical equation for the combustion of propane how many moles of c₃h₈ are present at the start of the reaction? moles how much of the excess reactant would be left over? moles what is the limiting reactant? need help determining the moles of each reactant? want some help finding the limiting reactant? how much of the excess reactant is left over?

Explanation:

Step1: Calculate moles of \(C_3H_8\)

Use formula \(n=\frac{m}{M}\), molar mass of \(C_3H_8(M)=(3\times12)+(8\times1)=44\ g/mol\).
\(n_{C_3H_8}=\frac{153.8\ g}{44\ g/mol}\approx3.5\ mol\)

Step2: Calculate moles of \(O_2\)

Molar mass of \(O_2(M) = 32\ g/mol\).
\(n_{O_2}=\frac{352\ g}{32\ g/mol}= 11\ mol\)

Step3: Determine limiting reactant

From balanced equation \(C_3H_8 + 5O_2
ightarrow3CO_2 + 4H_2O\), mole ratio \(C_3H_8:O_2 = 1:5\).
For \(3.5\ mol\ C_3H_8\), required \(O_2\) is \(3.5\times5 = 17.5\ mol\). But we have \(11\ mol\ O_2\).
For \(11\ mol\ O_2\), required \(C_3H_8\) is \(\frac{11}{5}=2.2\ mol\).
Since \(O_2\) is completely consumed first, \(O_2\) is limiting reactant.

Step4: Calculate excess \(C_3H_8\)

Used \(C_3H_8 = 2.2\ mol\). Initial \(C_3H_8=3.5\ mol\).
Excess \(C_3H_8=3.5 - 2.2=1.3\ mol\)

Answer:

• Moles of \(C_3H_8\) at start: \(3.5\) moles
• Limiting reactant: \(O_2\)
• Excess reactant left: \(1.3\) moles