Answer :
Certainly! Here is a detailed, step-by-step solution to each part of the question based on stoichiometric calculations from the given chemical reaction:
The given balanced chemical reaction is:
[tex]\[ 4 NH_3 + 5 O_2 \rightarrow 6 H_2O + 4 NO \][/tex]
1. What mass of NO is produced when 2 mol of [tex]\( NH_3 \)[/tex] reacts with excess [tex]\( O_2 \)[/tex]?
- According to the balanced equation: [tex]\( 4 \)[/tex] moles of [tex]\( NH_3 \)[/tex] produce [tex]\( 4 \)[/tex] moles of [tex]\( NO \)[/tex].
- Therefore, [tex]\( 2 \)[/tex] moles of [tex]\( NH_3 \)[/tex] (which is half of 4 moles) will produce [tex]\( 2 \)[/tex] moles of [tex]\( NO \)[/tex].
- The molar mass of [tex]\( NO \)[/tex] is [tex]\( 30 \)[/tex] g/mol.
- The mass of [tex]\( NO \)[/tex] produced can be calculated using the formula:
[tex]\[ \text{Mass of NO} = \text{Moles of NO} \times \text{Molar mass of NO} \][/tex]
[tex]\[ \text{Mass of NO} = 2 \text{ mol} \times 30 \text{ g/mol} = 60 \text{ g} \][/tex]
So, the mass of NO produced is [tex]\( 60 \text{ g} \)[/tex].
2. What mass of [tex]\( H_2O \)[/tex] is produced when 4 mol of [tex]\( NH_3 \)[/tex] reacts with excess [tex]\( O_2 \)[/tex]?
- According to the balanced equation: [tex]\( 4 \)[/tex] moles of [tex]\( NH_3 \)[/tex] produce [tex]\( 6 \)[/tex] moles of [tex]\( H_2O \)[/tex].
- Therefore, [tex]\( 4 \)[/tex] moles of [tex]\( NH_3 \)[/tex] will produce [tex]\( 6 \)[/tex] moles of [tex]\( H_2O \)[/tex].
- The molar mass of [tex]\( H_2O \)[/tex] is [tex]\( 18 \)[/tex] g/mol.
- The mass of [tex]\( H_2O \)[/tex] produced can be calculated using the formula:
[tex]\[ \text{Mass of H}_2\text{O} = \text{Moles of H}_2\text{O} \times \text{Molar mass of H}_2\text{O} \][/tex]
[tex]\[ \text{Mass of H}_2\text{O} = 6 \text{ mol} \times 18 \text{ g/mol} = 108 \text{ g} \][/tex]
So, the mass of [tex]\( H_2O \)[/tex] produced is [tex]\( 108 \text{ g} \)[/tex].
3. What mass of [tex]\( O_2 \)[/tex] must react to form [tex]\( 2.5 \)[/tex] mol of NO?
- According to the balanced equation: [tex]\( 4 \)[/tex] moles of [tex]\( NO \)[/tex] are produced from [tex]\( 5 \)[/tex] moles of [tex]\( O_2 \)[/tex].
- Therefore, [tex]\( 2.5 \)[/tex] moles of [tex]\( NO \)[/tex] will need:
[tex]\[ \frac{2.5 \text{ mol NO}}{4 \text{ mol NO}} \times 5 \text{ mol O}_2 = 3.125 \text{ mol O}_2 \][/tex]
- The molar mass of [tex]\( O_2 \)[/tex] is [tex]\( 32 \)[/tex] g/mol.
- The mass of [tex]\( O_2 \)[/tex] required can be calculated using the formula:
[tex]\[ \text{Mass of O}_2 = \text{Moles of O}_2 \times \text{Molar mass of O}_2 \][/tex]
[tex]\[ \text{Mass of O}_2 = 3.125 \text{ mol} \times 32 \text{ g/mol} = 100 \text{ g} \][/tex]
So, the mass of [tex]\( O_2 \)[/tex] required is [tex]\( 100 \text{ g} \)[/tex].
To summarize:
1. The mass of NO produced when 2 mol of [tex]\( NH_3 \)[/tex] reacts with excess [tex]\( O_2 \)[/tex] is [tex]\( 60 \text{ g} \)[/tex].
2. The mass of [tex]\( H_2O \)[/tex] produced when 4 mol of [tex]\( NH_3 \)[/tex] reacts with excess [tex]\( O_2 \)[/tex] is [tex]\( 108 \text{ g} \)[/tex].
3. The mass of [tex]\( O_2 \)[/tex] required to form [tex]\( 2.5 \)[/tex] mol of NO is [tex]\( 100 \text{ g} \)[/tex].
The given balanced chemical reaction is:
[tex]\[ 4 NH_3 + 5 O_2 \rightarrow 6 H_2O + 4 NO \][/tex]
1. What mass of NO is produced when 2 mol of [tex]\( NH_3 \)[/tex] reacts with excess [tex]\( O_2 \)[/tex]?
- According to the balanced equation: [tex]\( 4 \)[/tex] moles of [tex]\( NH_3 \)[/tex] produce [tex]\( 4 \)[/tex] moles of [tex]\( NO \)[/tex].
- Therefore, [tex]\( 2 \)[/tex] moles of [tex]\( NH_3 \)[/tex] (which is half of 4 moles) will produce [tex]\( 2 \)[/tex] moles of [tex]\( NO \)[/tex].
- The molar mass of [tex]\( NO \)[/tex] is [tex]\( 30 \)[/tex] g/mol.
- The mass of [tex]\( NO \)[/tex] produced can be calculated using the formula:
[tex]\[ \text{Mass of NO} = \text{Moles of NO} \times \text{Molar mass of NO} \][/tex]
[tex]\[ \text{Mass of NO} = 2 \text{ mol} \times 30 \text{ g/mol} = 60 \text{ g} \][/tex]
So, the mass of NO produced is [tex]\( 60 \text{ g} \)[/tex].
2. What mass of [tex]\( H_2O \)[/tex] is produced when 4 mol of [tex]\( NH_3 \)[/tex] reacts with excess [tex]\( O_2 \)[/tex]?
- According to the balanced equation: [tex]\( 4 \)[/tex] moles of [tex]\( NH_3 \)[/tex] produce [tex]\( 6 \)[/tex] moles of [tex]\( H_2O \)[/tex].
- Therefore, [tex]\( 4 \)[/tex] moles of [tex]\( NH_3 \)[/tex] will produce [tex]\( 6 \)[/tex] moles of [tex]\( H_2O \)[/tex].
- The molar mass of [tex]\( H_2O \)[/tex] is [tex]\( 18 \)[/tex] g/mol.
- The mass of [tex]\( H_2O \)[/tex] produced can be calculated using the formula:
[tex]\[ \text{Mass of H}_2\text{O} = \text{Moles of H}_2\text{O} \times \text{Molar mass of H}_2\text{O} \][/tex]
[tex]\[ \text{Mass of H}_2\text{O} = 6 \text{ mol} \times 18 \text{ g/mol} = 108 \text{ g} \][/tex]
So, the mass of [tex]\( H_2O \)[/tex] produced is [tex]\( 108 \text{ g} \)[/tex].
3. What mass of [tex]\( O_2 \)[/tex] must react to form [tex]\( 2.5 \)[/tex] mol of NO?
- According to the balanced equation: [tex]\( 4 \)[/tex] moles of [tex]\( NO \)[/tex] are produced from [tex]\( 5 \)[/tex] moles of [tex]\( O_2 \)[/tex].
- Therefore, [tex]\( 2.5 \)[/tex] moles of [tex]\( NO \)[/tex] will need:
[tex]\[ \frac{2.5 \text{ mol NO}}{4 \text{ mol NO}} \times 5 \text{ mol O}_2 = 3.125 \text{ mol O}_2 \][/tex]
- The molar mass of [tex]\( O_2 \)[/tex] is [tex]\( 32 \)[/tex] g/mol.
- The mass of [tex]\( O_2 \)[/tex] required can be calculated using the formula:
[tex]\[ \text{Mass of O}_2 = \text{Moles of O}_2 \times \text{Molar mass of O}_2 \][/tex]
[tex]\[ \text{Mass of O}_2 = 3.125 \text{ mol} \times 32 \text{ g/mol} = 100 \text{ g} \][/tex]
So, the mass of [tex]\( O_2 \)[/tex] required is [tex]\( 100 \text{ g} \)[/tex].
To summarize:
1. The mass of NO produced when 2 mol of [tex]\( NH_3 \)[/tex] reacts with excess [tex]\( O_2 \)[/tex] is [tex]\( 60 \text{ g} \)[/tex].
2. The mass of [tex]\( H_2O \)[/tex] produced when 4 mol of [tex]\( NH_3 \)[/tex] reacts with excess [tex]\( O_2 \)[/tex] is [tex]\( 108 \text{ g} \)[/tex].
3. The mass of [tex]\( O_2 \)[/tex] required to form [tex]\( 2.5 \)[/tex] mol of NO is [tex]\( 100 \text{ g} \)[/tex].
