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------------------------------------------------ Consider the following reaction:

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

How many grams of water vapor can be produced if you begin with 37.1 grams of each reactant?

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.