Question

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

4 NH3 + 3 O2  4 NO + 6 H2O

Answer 1

To find out how many grams of NH₃ (ammonia) are needed to form 28.4 moles of H₂O (water), we can first look at the balanced chemical reaction:

\[ 4 \, \text{NH}_3 + 3 \, \text{O}_2 \rightarrow 4 \, \text{NO} + 6 \, \text{H}_2O \]

From the balanced equation, we can see the stoichiometric relationship between NH₃ and H₂O. Specifically, for every 4 moles of NH₃, 6 moles of H₂O are produced.

Next, we can set up a proportion to find out how many moles of NH₃ are needed to produce 28.4 moles of H₂O:

\[
\frac{4 \, \text{NH}_3}{6 \, \text{H}_2O} = \frac{x \, \text{NH}_3}{28.4 \, \text{H}_2O}
\]

Where \( x \) is the number of moles of NH₃ needed. Cross-multiplying gives us:

\[
6x = 4 \times 28.4
\]

Calculating the right side:

\[
6x = 113.6
\]

Now, divide both sides by 6 to solve for \( x \):

\[
x = \frac{113.6}{6} \approx 18.9333 \, \text{moles of NH}_3
\]

Now that we have the number of moles of NH₃ needed, we can convert that to grams. The molar mass of NH₃ is approximately \( 14.01 \, \text{(N)} + 3 \times 1.01 \, \text{(H)} = 17.04 \, \text{g/mol} \).

Now we can calculate the mass of NH₃ needed:

\[
\text{Mass of NH}_3 = \text{moles} \times \text{molar mass} = 18.9333 \, \text{moles} \times 17.04 \, \text{g/mol}
\]

\[
\text{Mass of NH}_3 \approx 322.36 \, \text{grams}
\]

Therefore, you need approximately **322.36 grams of NH₃** to produce 28.4 moles of H₂O.