Answer :
Certainly! Let's solve this problem step by step.
We are given the reaction:
[tex]\[ C_3H_8 + 5O_2 \rightarrow 3CO_2 + 4H_2O \][/tex]
And we are asked how many grams of water vapor ([tex]\( H_2O \)[/tex]) can be produced if we start with 37.1 grams of each reactant ([tex]\( C_3H_8 \)[/tex] and [tex]\( O_2 \)[/tex]).
### Step 1: Calculate Moles of Reactants
1. Calculate moles of [tex]\( C_3H_8 \)[/tex]:
The molar mass of [tex]\( C_3H_8 \)[/tex] (propane) is approximately 44.10 grams per mole.
[tex]\[
\text{Moles of } C_3H_8 = \frac{\text{mass of } C_3H_8}{\text{molar mass of } C_3H_8} = \frac{37.1 \, \text{g}}{44.10 \, \text{g/mol}} \approx 0.841 \, \text{moles}
\][/tex]
2. Calculate moles of [tex]\( O_2 \)[/tex]:
The molar mass of [tex]\( O_2 \)[/tex] is approximately 32.00 grams per mole.
[tex]\[
\text{Moles of } O_2 = \frac{\text{mass of } O_2}{\text{molar mass of } O_2} = \frac{37.1 \, \text{g}}{32.00 \, \text{g/mol}} \approx 1.159 \, \text{moles}
\][/tex]
### Step 2: Identify the Limiting Reactant
To find the limiting reactant, we use the stoichiometry of the reaction. From the balanced equation, 1 mole of [tex]\( C_3H_8 \)[/tex] reacts with 5 moles of [tex]\( O_2 \)[/tex].
- Check for [tex]\( C_3H_8 \)[/tex]:
[tex]\[
\frac{0.841 \, \text{moles of } C_3H_8}{1} = 0.841
\][/tex]
- Check for [tex]\( O_2 \)[/tex]:
[tex]\[
\frac{1.159 \, \text{moles of } O_2}{5} = 0.232
\][/tex]
The limiting reactant is the one with the smaller ratio, which is [tex]\( O_2 \)[/tex].
### Step 3: Calculate Moles of [tex]\( H_2O \)[/tex] Produced
Since [tex]\( O_2 \)[/tex] is the limiting reactant, we use it to determine how much water is produced.
According to the equation, 5 moles of [tex]\( O_2 \)[/tex] produce 4 moles of [tex]\( H_2O \)[/tex].
[tex]\[
\text{Moles of } H_2O = 0.232 \times 4 = 0.928 \, \text{moles}
\][/tex]
### Step 4: Convert Moles of [tex]\( H_2O \)[/tex] to Grams
The molar mass of [tex]\( H_2O \)[/tex] is approximately 18.02 grams per mole.
[tex]\[
\text{Mass of } H_2O = \text{moles of } H_2O \times \text{molar mass of } H_2O = 0.928 \, \text{moles} \times 18.02 \, \text{g/mol} \approx 16.71 \, \text{grams}
\][/tex]
Therefore, approximately 16.71 grams of water vapor can be produced from 37.1 grams of each reactant.
We are given the reaction:
[tex]\[ C_3H_8 + 5O_2 \rightarrow 3CO_2 + 4H_2O \][/tex]
And we are asked how many grams of water vapor ([tex]\( H_2O \)[/tex]) can be produced if we start with 37.1 grams of each reactant ([tex]\( C_3H_8 \)[/tex] and [tex]\( O_2 \)[/tex]).
### Step 1: Calculate Moles of Reactants
1. Calculate moles of [tex]\( C_3H_8 \)[/tex]:
The molar mass of [tex]\( C_3H_8 \)[/tex] (propane) is approximately 44.10 grams per mole.
[tex]\[
\text{Moles of } C_3H_8 = \frac{\text{mass of } C_3H_8}{\text{molar mass of } C_3H_8} = \frac{37.1 \, \text{g}}{44.10 \, \text{g/mol}} \approx 0.841 \, \text{moles}
\][/tex]
2. Calculate moles of [tex]\( O_2 \)[/tex]:
The molar mass of [tex]\( O_2 \)[/tex] is approximately 32.00 grams per mole.
[tex]\[
\text{Moles of } O_2 = \frac{\text{mass of } O_2}{\text{molar mass of } O_2} = \frac{37.1 \, \text{g}}{32.00 \, \text{g/mol}} \approx 1.159 \, \text{moles}
\][/tex]
### Step 2: Identify the Limiting Reactant
To find the limiting reactant, we use the stoichiometry of the reaction. From the balanced equation, 1 mole of [tex]\( C_3H_8 \)[/tex] reacts with 5 moles of [tex]\( O_2 \)[/tex].
- Check for [tex]\( C_3H_8 \)[/tex]:
[tex]\[
\frac{0.841 \, \text{moles of } C_3H_8}{1} = 0.841
\][/tex]
- Check for [tex]\( O_2 \)[/tex]:
[tex]\[
\frac{1.159 \, \text{moles of } O_2}{5} = 0.232
\][/tex]
The limiting reactant is the one with the smaller ratio, which is [tex]\( O_2 \)[/tex].
### Step 3: Calculate Moles of [tex]\( H_2O \)[/tex] Produced
Since [tex]\( O_2 \)[/tex] is the limiting reactant, we use it to determine how much water is produced.
According to the equation, 5 moles of [tex]\( O_2 \)[/tex] produce 4 moles of [tex]\( H_2O \)[/tex].
[tex]\[
\text{Moles of } H_2O = 0.232 \times 4 = 0.928 \, \text{moles}
\][/tex]
### Step 4: Convert Moles of [tex]\( H_2O \)[/tex] to Grams
The molar mass of [tex]\( H_2O \)[/tex] is approximately 18.02 grams per mole.
[tex]\[
\text{Mass of } H_2O = \text{moles of } H_2O \times \text{molar mass of } H_2O = 0.928 \, \text{moles} \times 18.02 \, \text{g/mol} \approx 16.71 \, \text{grams}
\][/tex]
Therefore, approximately 16.71 grams of water vapor can be produced from 37.1 grams of each reactant.
