Alright, let's determine the percent yield of water produced in the given reaction step-by-step. Here are the given values and known constants:
1. Mass of hydrogen ([tex]\( H_2 \)[/tex]): 68.3 g
2. Mass of oxygen ([tex]\( O_2 \)[/tex]): 85.4 g
3. Mass of water collected ([tex]\( H_2O \)[/tex]): 86.4 g
Molar masses (g/mol) are:
1. Molar mass of [tex]\( H_2 \)[/tex]: 2.016 g/mol
2. Molar mass of [tex]\( O_2 \)[/tex]: 32.00 g/mol
3. Molar mass of [tex]\( H_2O \)[/tex]: 18.015 g/mol
We need to follow these steps:
### Step 1: Calculate the moles of hydrogen and oxygen
First, calculate the moles of hydrogen [tex]\( H_2 \)[/tex]:
[tex]\[
\text{moles of } H_2 = \frac{\text{mass of } H_2}{\text{molar mass of } H_2} = \frac{68.3 \text{ g}}{2.016 \text{ g/mol}} \approx 33.879 \text{ moles}
\][/tex]
Next, calculate the moles of oxygen [tex]\( O_2 \)[/tex]:
[tex]\[
\text{moles of } O_2 = \frac{\text{mass of } O_2}{\text{molar mass of } O_2} = \frac{85.4 \text{ g}}{32.00 \text{ g/mol}} \approx 2.669 \text{ moles}
\][/tex]
### Step 2: Determine the stoichiometric relationship
The balanced equation for the reaction is:
[tex]\[ 2 H_2 + O_2 \rightarrow 2 H_2O \][/tex]
This tells us that 2 moles of [tex]\( H_2 \)[/tex] react with 1 mole of [tex]\( O_2 \)[/tex] to produce 2 moles of [tex]\( H_2O \)[/tex].
### Step 3: Identify the limiting reactant
Calculate the ideal moles of [tex]\( H_2O \)[/tex] produced using hydrogen and oxygen:
For hydrogen:
[tex]\[
\text{From 33.879 moles of } H_2, \text{ we can get:} \quad \left( \frac{2 \text{ moles } H_2O}{2 \text{ moles } H_2} \right) \times 33.879 = 33.879 \text{ moles } H_2O
\][/tex]
For oxygen:
[tex]\[
\text{From 2.669 moles of } O_2, \text{ we can get:} \quad \left( \frac{2 \text{ moles } H_2O}{1 \text{ mole } O_2} \right) \times 2.669 = 5.339 \text{ moles } H_2O
\][/tex]
The limiting reactant is [tex]\( O_2 \)[/tex] because it produces fewer moles of [tex]\( H_2O \)[/tex].
### Step 4: Calculate the theoretical yield of water
The moles of water produced ideally are determined by the limiting reactant:
[tex]\[
\text{Moles of } H_2O \text{ produced} = 5.339 \text{ moles}
\][/tex]
Converting this to mass, we get:
[tex]\[
\text{Mass of } H_2O = \text{moles of } H_2O \times \text{molar mass of } H_2O = 5.339 \times 18.015 \approx 96.155 \text{ g}
\][/tex]
### Step 5: Calculate the percent yield
Finally, the percent yield is:
[tex]\[
\text{Percent yield} = \left( \frac{\text{actual yield}}{\text{theoretical yield}} \right) \times 100 = \left( \frac{86.4 \text{ g}}{96.155 \text{ g}} \right) \times 100 \approx 89.85\%
\][/tex]
Thus, the percent yield of water produced in this reaction is approximately 89.85%.
