Question

missed this? hatch kcc: reaction stoichiometry. read section 8.3. you can click on the review link to access the section in your etext.
for this 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.

$$ mathrm { c } _ { 3 } mathrm { h } _ { 8 } ( g ) + 5 mathrm { o } _ { 2 } ( g ) ightarrow 3 mathrm { co } _ { 2 } ( g ) + 4 mathrm { h } _ { 2 } mathrm { o } ( g ) $$

part f
$$ 4.0 mathrm { mol } mathrm { o } _ { 2 } $$
express your answer using two significant figures.
part g
$$ 0.0551 mathrm { mol } mathrm { o } _ { 2 } $$
express your answer using three significant figures.
part h
$$ 0.0551 mathrm { mol } mathrm { o } _ { 2 } $$
express your answer using three significant figures.

Explanation:

Step1: Determine mole ratio

From the balanced equation \(C_3H_8(g)+5O_2(g)\to3CO_2(g)+4H_2O(g)\), the mole ratio of \(O_2\) to \(H_2O\) is \(5:4\).

Step2: Calculate moles of \(H_2O\) for Part F

Given \(n(O_2) = 4.0\space mol\). Using the mole ratio \(\frac{n(H_2O)}{n(O_2)}=\frac{4}{5}\), so \(n(H_2O)=\frac{4}{5}\times n(O_2)\). Substitute \(n(O_2) = 4.0\space mol\), \(n(H_2O)=\frac{4}{5}\times4.0\space mol = 3.2\space mol\)

Step3: Determine mole ratio for \(O_2\) and \(CO_2\)

From the balanced equation, the mole ratio of \(O_2\) to \(CO_2\) is \(5:3\)

Step4: Calculate moles of \(CO_2\) for Part G

Given \(n(O_2)=0.0551\space mol\). Using the mole ratio \(\frac{n(CO_2)}{n(O_2)}=\frac{3}{5}\), so \(n(CO_2)=\frac{3}{5}\times n(O_2)\). Substitute \(n(O_2) = 0.0551\space mol\), \(n(CO_2)=\frac{3}{5}\times0.0551\space mol=0.0331\space mol\)

Answer:

Part F: \(3.2\space mol\space H_2O\)
Part G: \(0.0331\space mol\space CO_2\)