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Sagot :
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].
### 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].
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