The chemical equation below shows the combustion of propane [tex] \left( C _3 H _8\right) [/tex].

[tex]\[ C_3H_8 + 5O_2 \rightarrow 3CO_2 + 4H_2O \][/tex]

The molar mass of oxygen gas [tex] \left( O_2 \right) [/tex] is [tex] 32.00 \, \text{g/mol} [/tex]. The molar mass of [tex] C_3H_8 [/tex] is [tex] 44.1 \, \text{g/mol} [/tex].

What mass of [tex] O_2 [/tex], in grams, is required to completely react with [tex] 0.025 \, \text{g} \, C_3H_8 [/tex]?

A. 0.018 grams
B. 0.034 grams
C. 0.045 grams
D. 0.091 grams



Answer :

To determine the mass of [tex]\(O_2\)[/tex] needed to completely react with [tex]\(0.025 \text{ g}\)[/tex] of [tex]\(C_3H_8\)[/tex], let's follow through a detailed, step-by-step solution.

### Step 1: Calculate the Moles of [tex]\(C_3H_8\)[/tex]

First, we determine the number of moles of [tex]\(C_3H_8\)[/tex] we have using the given mass and the molar mass.

[tex]\[ \text{Molar mass of } C_3H_8 = 44.1 \, \text{g/mol} \][/tex]
[tex]\[ \text{Mass of } C_3H_8 = 0.025 \, \text{g} \][/tex]
[tex]\[ \text{Moles of } C_3H_8 = \frac{\text{Mass of } C_3H_8}{\text{Molar mass of } C_3H_8} = \frac{0.025 \, \text{g}}{44.1 \, \text{g/mol}} \approx 0.000566893 \, \text{mol} \][/tex]

### Step 2: Use Stoichiometry to Determine the Moles of [tex]\(O_2\)[/tex] Needed

The balanced chemical equation for the combustion of propane is:
[tex]\[ C_3H_8 + 5O_2 \rightarrow 3CO_2 + 4H_2O \][/tex]

From this equation, we see that 1 mole of [tex]\(C_3H_8\)[/tex] reacts with 5 moles of [tex]\(O_2\)[/tex]. Therefore, we can find the moles of [tex]\(O_2\)[/tex] needed to react with our [tex]\(C_3H_8\)[/tex]:

[tex]\[ \text{Moles of } O_2 \, \text{needed} = \text{Moles of } C_3H_8 \times 5 = 0.000566893 \, \text{mol} \times 5 \approx 0.002834467 \, \text{mol} \][/tex]

### Step 3: Calculate the Mass of [tex]\(O_2\)[/tex] Needed

Finally, we determine the mass of [tex]\(O_2\)[/tex] required by using the moles of [tex]\(O_2\)[/tex] needed and the molar mass of [tex]\(O_2\)[/tex].

[tex]\[ \text{Molar mass of } O_2 = 32.00 \, \text{g/mol} \][/tex]
[tex]\[ \text{Mass of } O_2 = \text{Moles of } O_2 \, \text{needed} \times \text{Molar mass of } O_2 = 0.002834467 \, \text{mol} \times 32.00 \, \text{g/mol} \approx 0.0907029 \, \text{g} \][/tex]

### Step 4: Choose the Closest Answer from Given Options

Among the provided multiple-choice options: [tex]\(0.018 \, \text{grams}\)[/tex], [tex]\(0.034 \, \text{grams}\)[/tex], [tex]\(0.045 \, \text{grams}\)[/tex], and [tex]\(0.091 \, \text{grams}\)[/tex], the closest value to our calculated mass ([tex]\(0.0907029 \, \text{grams}\)[/tex]) is [tex]\(0.091 \, \text{grams}\)[/tex].

Therefore, the mass of [tex]\(O_2\)[/tex] required to completely react with [tex]\(0.025 \text{ g}\)[/tex] of [tex]\(C_3H_8\)[/tex] is [tex]\(\boxed{0.091 \, \text{grams}}\)[/tex].