College

If 39.1 g of [tex]$NO$[/tex] and 26.9 g of [tex]$O_2$[/tex] react together, what is the mass, in grams, of [tex]$NO_2$[/tex] that can be formed?

The balanced equation is:
[tex]2 NO(g) + O_2(g) \rightarrow 2 NO_2(g)[/tex]

Answer :

To determine the mass of NO₂ formed when 39.1 g of NO and 26.9 g of O₂ react, follow these steps:

1. Calculate Moles of Each Reactant:

- NO: The molar mass of NO is 30.01 g/mol.
[tex]\[
\text{Moles of NO} = \frac{39.1 \, \text{g}}{30.01 \, \text{g/mol}} \approx 1.303 \, \text{moles}
\][/tex]

- O₂: The molar mass of O₂ is 32.00 g/mol.
[tex]\[
\text{Moles of O₂} = \frac{26.9 \, \text{g}}{32.00 \, \text{g/mol}} \approx 0.841 \, \text{moles}
\][/tex]

2. Identify the Limiting Reactant:

The balanced equation is:
[tex]\[
2 \, \text{NO} + \text{O}_2 \rightarrow 2 \, \text{NO}_2
\][/tex]
According to the equation, 2 moles of NO react with 1 mole of O₂.

- Moles of NO needed for available O₂:
[tex]\[
2 \times 0.841 \, \text{moles of O}_2 = 1.682 \, \text{moles of NO}
\][/tex]
- Since we have only 1.303 moles of NO, NO is the limiting reactant.

3. Calculate Moles of NO₂ Formed:

From the balanced equation, 2 moles of NO produce 2 moles of NO₂. Therefore:
[tex]\[
\text{Moles of NO}_2 = 1.303 \, \text{moles of NO}
\][/tex]

4. Convert Moles of NO₂ to Mass:

The molar mass of NO₂ is 46.01 g/mol. Thus, the mass of NO₂ is:
[tex]\[
\text{Mass of NO}_2 = 1.303 \, \text{moles} \times 46.01 \, \text{g/mol} \approx 59.95 \, \text{g}
\][/tex]

Therefore, the mass of NO₂ formed is approximately 59.95 grams.

Other Questions