To determine how many grams of [tex]\( CO_2 \)[/tex] can be produced by reacting 1.088 grams of oxygen gas ([tex]\( O_2 \)[/tex]), follow these steps:
1. Find the Molar Masses:
- Molar mass of [tex]\( O_2 \)[/tex] is 32 grams per mole.
- Molar mass of [tex]\( CO_2 \)[/tex] is 44 grams per mole.
2. Calculate the Number of Moles of [tex]\( O_2 \)[/tex]:
[tex]\[
\text{Number of moles of } O_2 = \frac{\text{mass of } O_2}{\text{molar mass of } O_2}
\][/tex]
Given mass of [tex]\( O_2 \)[/tex] is 1.088 grams:
[tex]\[
\text{Number of moles of } O_2 = \frac{1.088 \text{ grams}}{32 \text{ g/mol}} \approx 0.034 \text{ moles}
\][/tex]
3. Determine the Stoichiometric Ratio:
From the balanced equation:
[tex]\[
C_3 H_8 + 5 O_2 \rightarrow 3 CO_2 + 4 H_2O
\][/tex]
There is a stoichiometric ratio of 5 moles of [tex]\( O_2 \)[/tex] to 3 moles of [tex]\( CO_2 \)[/tex]. Thus:
[tex]\[
\text{Ratio of moles of } O_2 \text{ to moles of } CO_2 = \frac{3}{5}
\][/tex]
4. Calculate the Number of Moles of [tex]\( CO_2 \)[/tex] Produced:
[tex]\[
\text{Number of moles of } CO_2 = \text{moles of } O_2 \times \frac{3}{5}
\][/tex]
Substituting the number of moles of [tex]\( O_2 \)[/tex] calculated:
[tex]\[
\text{Number of moles of } CO_2 = 0.034 \text{ moles} \times \frac{3}{5} \approx 0.0204 \text{ moles}
\][/tex]
5. Calculate the Mass of [tex]\( CO_2 \)[/tex] Produced:
[tex]\[
\text{Mass of } CO_2 = \text{number of moles of } CO_2 \times \text{molar mass of } CO_2
\][/tex]
Given the molar mass of [tex]\( CO_2 \)[/tex] is 44 grams per mole:
[tex]\[
\text{Mass of } CO_2 = 0.0204 \text{ moles} \times 44 \text{ g/mol} \approx 0.8976 \text{ grams}
\][/tex]
In summary, reacting 1.088 grams of oxygen gas will produce approximately 0.8976 grams of [tex]\( CO_2 \)[/tex].
