Question

What mass of ammonia, NH3, is necessary to react with 2.1 moles of oxygen, given 4NH3 + 7O2 --> 6H2O + 4NO2?
Responses

20.4 g
20.4 g

9.71 g
9.71 g

4.46 g
4.46 g

143 g

Answer 1

To find the mass of ammonia (NH₃) needed to react with 2.1 moles of oxygen (O₂) according to the reaction:

\[ 4 \text{NH}_3 + 7 \text{O}_2 \rightarrow 6 \text{H}_2\text{O} + 4 \text{NO}_2 \]

we first need to determine the stoichiometric ratio between ammonia and oxygen. From the balanced equation, we can see that:

- 4 moles of NH₃ react with 7 moles of O₂.

We can set up a ratio to find the number of moles of NH₃ needed to react with 2.1 moles of O₂:

\[
\frac{4 \text{ moles of NH}_3}{7 \text{ moles of O}_2} = \frac{x \text{ moles of NH}_3}{2.1 \text{ moles of O}_2}
\]

Now we can solve for \( x \):

\[
x = 2.1 \text{ moles O}_2 \times \frac{4 \text{ moles NH}_3}{7 \text{ moles O}_2} = \frac{8.4}{7} \text{ moles NH}_3 \approx 1.2 \text{ moles NH}_3
\]

Next, we need to convert the amount of NH₃ in moles to grams. The molar mass of NH₃ can be calculated as follows:

- N: 14.01 g/mol
- H: 1.01 g/mol (3 H atoms)

Calculating the molar mass of NH₃:

\[
\text{Molar mass of NH}_3 = 14.01 + (3 \times 1.01) = 14.01 + 3.03 = 17.04 \, \text{g/mol}
\]

Now, we convert moles of NH₃ to grams:

\[
\text{Mass of NH}_3 = 1.2 \text{ moles} \times 17.04 \text{ g/mol} \approx 20.448 \text{ g}
\]

Rounding this to three significant figures gives approximately 20.4 g.

Therefore, the mass of ammonia necessary to react with 2.1 moles of oxygen is:

**20.4 g**