Question

missed this? watch kcv reaction stoichiometry, read section 8.3. you can click on the review link to access the section in your etext. for the reaction shown, calculate how many moles of each product form when the given amount of each reactant completely reacts. assume that there is more than enough of the other reactant. ( c_{3}h_{8}(g)+5o_{2}(g)
ightarrow3co_{2}(g)+4h_{2}o(g) ) part a ( 4.0 mol c_{3}h_{8} ) express your answer using two significant figures. part b ( 4.0 mol c_{3}h_{8} ) express your answer using two significant figures. part c ( 0.0551 mol c_{3}h_{8} ) express your answer using three significant figures.

Explanation:

Step1: Determine mole ratios

From the balanced equation \(C_{3}H_{8}(g)+5O_{2}(g)\to3CO_{2}(g) + 4H_{2}O(g)\), the mole ratio of \(C_{3}H_{8}\) to \(CO_{2}\) is \(1:3\), and the mole ratio of \(C_{3}H_{8}\) to \(H_{2}O\) is \(1:4\).

Step2: Calculate moles of \(CO_{2}\) for Part A

Given \(n(C_{3}H_{8}) = 4.0\space mol\). Using the mole ratio \(\frac{n(CO_{2})}{n(C_{3}H_{8})}=\frac{3}{1}\), so \(n(CO_{2})=4.0\space mol\times3 = 12\space mol\)

Step3: Calculate moles of \(H_{2}O\) for Part B

Using the mole ratio \(\frac{n(H_{2}O)}{n(C_{3}H_{8})}=\frac{4}{1}\), so \(n(H_{2}O)=4.0\space mol\times4 = 16\space mol\)

Answer:

Part A: \(12\space mol\space CO_{2}\)
Part B: \(16\space mol\space H_{2}O\)