# Question #a1c07

Jan 27, 2018

$= 8.97 \times {10}^{22} {C}_{3} {H}_{8} \text{ molecules}$

#### Explanation:

1. Write the balanced equation

${C}_{3} {H}_{8} \left(g\right) + 5 {O}_{2} \left(g\right) \to 3 C {O}_{2} \left(g\right) + 4 {H}_{2} O \left(g\right)$

2. Assuming that the $C {O}_{2}$ gas produced from the reaction behaved ideally; thus, the relationship that $\text{1mol of any ideal gas occupies a volume of 22.4L}$ is the equivalence statement needed to find the unknown variable and can be interpreted as:

$\text{1mol of any ideal gas}$$\equiv 22.4 L$ $\text{of any ideal gas}$

3. Given the desired volume of $C {O}_{2}$; that is, 10.0L and the relationship described above, it can be deduced that the number of moles $\left(\eta\right)$:

$\eta = 10.0 \cancel{L \cdot C {O}_{2}} \times \frac{1 m o l C {O}_{2}}{22.4 \cancel{L \cdot C {O}_{2}}} = 0.446 m o l C {O}_{2}$

4. Now, find the number of moles of ${C}_{3} {H}_{8}$. To convert $\eta C {O}_{2}$ to $\eta {C}_{3} {H}_{8}$, refer to the balanced equation for the mole ratio; i.e.,

$= 0.446 \cancel{m o l C {O}_{2}} \times \frac{1 m o l {C}_{3} {H}_{8}}{3 \cancel{m o l C {O}_{2}}}$
$= 0.149 m o l {C}_{3} {H}_{8}$

5. Knowing the number of moles of ${C}_{3} {H}_{8}$, the number of molecules can be obtained through the relationship:

$\text{1mol of } {C}_{3} {H}_{8}$ $\equiv 6.02 \times {10}^{23} {C}_{3} {H}_{8} \text{ molecules}$

6. Now, using the preceding equivalence statement where a suitable conversion factor is obtainable, the number of molecules of ${C}_{3} {H}_{8}$ is;

$= 0.149 \cancel{m o l {C}_{3} {H}_{8}} \times \frac{6.02 \times {10}^{23} {C}_{3} {H}_{8} \text{ molecules}}{1 \cancel{m o l {C}_{3} {H}_{8}}}$
$= 0.897 \times {10}^{23} {C}_{3} {H}_{8} \text{ molecules}$
$= 8.97 \times {10}^{22} {C}_{3} {H}_{8} \text{ molecules}$