To determine the mass of ammonia (NH3) necessary 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 find the stoichiometric relationship between NH3 and O2 from the balanced equation.
From the balanced equation, we can see the mole ratio of NH3 to O2 is:
\[ \frac{4 \text{ moles of NH}_3}{7 \text{ moles of O}_2} \]
This means that for every 7 moles of O2, 4 moles of NH3 are required.
Now, we need to calculate how many moles of NH3 are needed for 2.1 moles of O2:
Using the mole ratio:
\[ \text{Moles of NH}_3 = \frac{4 \text{ moles of NH}_3}{7 \text{ moles of O}_2} \times 2.1 \text{ moles of O}_2 \]
Calculating:
\[ \text{Moles of NH}_3 = \frac{4}{7} \times 2.1 = \frac{4 \times 2.1}{7} \approx \frac{8.4}{7} \approx 1.2 \text{ moles of NH}_3 \]
Now, we need to convert moles of NH3 to grams. The molar mass of NH3 is:
\[ \text{Molar mass of NH}_3 = (1 \times 14.01) + (3 \times 1.008) = 14.01 + 3.024 = 17.034 \text{ g/mol} \]
Now we can calculate the mass of NH3 needed:
\[ \text{Mass of NH}_3 = \text{moles of NH}_3 \times \text{molar mass of NH}_3 \]
\[ \text{Mass of NH}_3 = 1.2 \text{ moles} \times 17.034 \text{ g/mol} \approx 20.441 \text{ g} \]
Therefore, the mass of ammonia necessary to react with 2.1 moles of oxygen is approximately 20.44 grams.
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