Question

How many grams of NH3 are needed to form 28.4 moles of H2O?

4 NH3 + 3 O2  4 NO + 6 H2O
Responses

6.67 g
6.67 g

322 g
322 g

0.3 g
0.3 g

88.75 g

Answer 1

To determine how many grams of NH3 (ammonia) are needed to form 28.4 moles of H2O (water), we need to start with the balanced chemical equation:

\[ 4 \text{NH}_3 + 3 \text{O}_2 \rightarrow 4 \text{NO} + 6 \text{H}_2\text{O} \]

From the equation, we can see the stoichiometric relationship between NH3 and H2O:

- 4 moles of NH3 produce 6 moles of H2O.

Next, we need to determine how many moles of NH3 are required to produce 28.4 moles of H2O. We can set up a proportion based on the stoichiometry:

\[
\frac{4 \text{ moles NH}_3}{6 \text{ moles H}_2\text{O}} = \frac{x \text{ moles NH}_3}{28.4 \text{ moles H}_2\text{O}}
\]

Now we can solve for x (the moles of NH3 needed):

\[
x = \frac{4 \text{ moles NH}_3}{6 \text{ moles H}_2\text{O}} \times 28.4 \text{ moles H}_2\text{O}
\]

Calculating this gives:

\[
x = \frac{4}{6} \times 28.4 = \frac{2}{3} \times 28.4 = 18.93 \text{ moles NH}_3
\]

Next, we need to convert moles of NH3 to grams. The molar mass of NH3 (ammonia) is approximately:

- N: 14 g/mol
- H: 1 g/mol × 3 = 3 g/mol

So, molar mass of NH3:

\[
14 + 3 = 17 \text{ g/mol}
\]

Now we can calculate the mass in grams:

\[
\text{Mass} = 18.93 \text{ moles} \times 17 \text{ g/mol} = 321.81 \text{ g}
\]

Rounding to two significant figures gives us approximately **322 g**.

Thus, the answer is:

**322 g**