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.
\\( \mathrm { c } _ { 3 } \mathrm { h } _ { 8 } ( g ) + 5 \mathrm { o } _ { 2 } ( g ) \
ightarrow 3 \mathrm { co } _ { 2 } ( g ) + 4 \mathrm { h } _ { 2 } \mathrm { o } ( g ) \\)
\\( \
u = 3.2 \mathrm { mol } \mathrm { h } _ { 2 } \mathrm { o } \\)
previous answers
correct
part g
\\( 0.0551 \mathrm { mol } \mathrm { o } _ { 2 } \\)
express your answer using three significant figures.
\\( \
u = 3.31 \times 10 ^ { - 2 } \mathrm { mol } \mathrm { co } _ { 2 } \\)
previous answers
correct
part h
\\( 0.0551 \mathrm { mol } \mathrm { o } _ { 2 } \\)
express your answer using three significant figures.
\\( \
u = \\) \\( \mathrm { mol } \mathrm { h } _ { 2 } \mathrm { o } \\)

Explanation:

Step1: Analyze the reaction stoichiometry

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

Step2: Set up the proportion

Let \(n\) be the number of moles of \(H_{2}O\). We have the proportion \(\frac{n}{0.0551\space mol}=\frac{4}{5}\).

Step3: Solve for \(n\)

Cross - multiply to get \(n=\frac{4\times0.0551\space mol}{5}\).

$$n = 0.0441\space mol$$

Answer:

\(0.0441\space mol\)