Sure, let's solve this problem step-by-step.
1. Write the given chemical reaction:
[tex]\[
C_3H_8(g) + 5 O_2(g) \rightarrow 3 CO_2(g) + 4 H_2O(g)
\][/tex]
2. Molar masses:
- Molar mass of [tex]\(O_2\)[/tex] (oxygen gas): [tex]\(32 \, \text{g/mol}\)[/tex]
- Molar mass of [tex]\(CO_2\)[/tex] (carbon dioxide): [tex]\(44 \, \text{g/mol}\)[/tex]
3. Determine the given mass of [tex]\(O_2\)[/tex]:
[tex]\(1.136 \, \text{g}\)[/tex]
4. Calculate the moles of [tex]\(O_2\)[/tex]:
The moles of a substance can be found using the formula:
[tex]\[
\text{Moles} = \frac{\text{Given mass}}{\text{Molar mass}}
\][/tex]
So, the moles of [tex]\(O_2\)[/tex] are:
[tex]\[
\text{Moles of } O_2 = \frac{1.136 \, \text{g}}{32 \, \text{g/mol}} = 0.0355 \, \text{mol}
\][/tex]
5. Use the stoichiometry of the reaction:
According to the balanced chemical equation, 5 moles of [tex]\(O_2\)[/tex] produce 3 moles of [tex]\(CO_2\)[/tex].
Therefore:
[tex]\[
\text{Moles of } CO_2 = \left( \frac{3}{5} \right) \times 0.0355 \, \text{mol} = 0.0213 \, \text{mol}
\][/tex]
6. Convert moles of [tex]\(CO_2\)[/tex] to grams:
Use the formula:
[tex]\[
\text{Mass} = \text{Moles} \times \text{Molar mass}
\][/tex]
So, the mass of [tex]\(CO_2\)[/tex] is:
[tex]\[
\text{Mass of } CO_2 = 0.0213 \, \text{mol} \times 44 \, \text{g/mol} = 0.9372 \, \text{g}
\][/tex]
Therefore, by reacting 1.136 grams of oxygen gas, you can produce approximately 0.9372 grams of carbon dioxide.
