Answer :
Sure! Let's solve this problem step-by-step:
We have the chemical reaction:
[tex]\[ 4 \, \text{NH}_3 + 5 \, \text{O}_2 \rightarrow 4 \, \text{NO} + 6 \, \text{H}_2\text{O} \][/tex]
We need to determine how many grams of water ([tex]\(\text{H}_2\text{O}\)[/tex]) will be formed when 22 grams of oxygen ([tex]\(\text{O}_2\)[/tex]) react.
### Step 1: Determine Molar Mass
1. Molar mass of [tex]\(\text{O}_2\)[/tex]:
- Oxygen ([tex]\(\text{O}\)[/tex]) has a molar mass of approximately 16.00 g/mol.
- Therefore, [tex]\(\text{O}_2\)[/tex] has a molar mass of [tex]\(32.00 \, \text{g/mol}\)[/tex].
2. Molar mass of [tex]\(\text{H}_2\text{O}\)[/tex]:
- Hydrogen ([tex]\(\text{H}\)[/tex]) has a molar mass of approximately 1.01 g/mol.
- Therefore, [tex]\(\text{H}_2\)[/tex] has a molar mass of [tex]\(2.02 \, \text{g/mol}\)[/tex].
- Add oxygen’s molar mass, so [tex]\(\text{H}_2\text{O}\)[/tex] has a molar mass of [tex]\(18.02 \, \text{g/mol}\)[/tex].
### Step 2: Calculate Moles of Oxygen
We have 22 grams of [tex]\(\text{O}_2\)[/tex].
- Calculate moles of [tex]\(\text{O}_2\)[/tex]:
[tex]\[
\text{Moles of } \text{O}_2 = \frac{\text{Mass of } \text{O}_2}{\text{Molar mass of } \text{O}_2} = \frac{22 \, \text{g}}{32.00 \, \text{g/mol}} = 0.6875 \, \text{moles}
\][/tex]
### Step 3: Use Stoichiometry to Find Moles of Water
According to the balanced equation, 5 moles of [tex]\(\text{O}_2\)[/tex] produce 6 moles of [tex]\(\text{H}_2\text{O}\)[/tex].
- Since we have 0.6875 moles of [tex]\(\text{O}_2\)[/tex], calculate the moles of [tex]\(\text{H}_2\text{O}\)[/tex]:
[tex]\[
\text{Moles of } \text{H}_2\text{O} = \left(\frac{6}{5}\right) \times 0.6875 = 0.825 \, \text{moles}
\][/tex]
### Step 4: Calculate Mass of Water
- Calculate the mass of [tex]\(\text{H}_2\text{O}\)[/tex]:
[tex]\[
\text{Mass of } \text{H}_2\text{O} = \text{Moles of } \text{H}_2\text{O} \times \text{Molar mass of } \text{H}_2\text{O} = 0.825 \times 18.02 \, \text{g/mol} = 14.87 \, \text{grams}
\][/tex]
Therefore, when 22 grams of oxygen react, 14.87 grams of water will be formed.
We have the chemical reaction:
[tex]\[ 4 \, \text{NH}_3 + 5 \, \text{O}_2 \rightarrow 4 \, \text{NO} + 6 \, \text{H}_2\text{O} \][/tex]
We need to determine how many grams of water ([tex]\(\text{H}_2\text{O}\)[/tex]) will be formed when 22 grams of oxygen ([tex]\(\text{O}_2\)[/tex]) react.
### Step 1: Determine Molar Mass
1. Molar mass of [tex]\(\text{O}_2\)[/tex]:
- Oxygen ([tex]\(\text{O}\)[/tex]) has a molar mass of approximately 16.00 g/mol.
- Therefore, [tex]\(\text{O}_2\)[/tex] has a molar mass of [tex]\(32.00 \, \text{g/mol}\)[/tex].
2. Molar mass of [tex]\(\text{H}_2\text{O}\)[/tex]:
- Hydrogen ([tex]\(\text{H}\)[/tex]) has a molar mass of approximately 1.01 g/mol.
- Therefore, [tex]\(\text{H}_2\)[/tex] has a molar mass of [tex]\(2.02 \, \text{g/mol}\)[/tex].
- Add oxygen’s molar mass, so [tex]\(\text{H}_2\text{O}\)[/tex] has a molar mass of [tex]\(18.02 \, \text{g/mol}\)[/tex].
### Step 2: Calculate Moles of Oxygen
We have 22 grams of [tex]\(\text{O}_2\)[/tex].
- Calculate moles of [tex]\(\text{O}_2\)[/tex]:
[tex]\[
\text{Moles of } \text{O}_2 = \frac{\text{Mass of } \text{O}_2}{\text{Molar mass of } \text{O}_2} = \frac{22 \, \text{g}}{32.00 \, \text{g/mol}} = 0.6875 \, \text{moles}
\][/tex]
### Step 3: Use Stoichiometry to Find Moles of Water
According to the balanced equation, 5 moles of [tex]\(\text{O}_2\)[/tex] produce 6 moles of [tex]\(\text{H}_2\text{O}\)[/tex].
- Since we have 0.6875 moles of [tex]\(\text{O}_2\)[/tex], calculate the moles of [tex]\(\text{H}_2\text{O}\)[/tex]:
[tex]\[
\text{Moles of } \text{H}_2\text{O} = \left(\frac{6}{5}\right) \times 0.6875 = 0.825 \, \text{moles}
\][/tex]
### Step 4: Calculate Mass of Water
- Calculate the mass of [tex]\(\text{H}_2\text{O}\)[/tex]:
[tex]\[
\text{Mass of } \text{H}_2\text{O} = \text{Moles of } \text{H}_2\text{O} \times \text{Molar mass of } \text{H}_2\text{O} = 0.825 \times 18.02 \, \text{g/mol} = 14.87 \, \text{grams}
\][/tex]
Therefore, when 22 grams of oxygen react, 14.87 grams of water will be formed.
