To determine the mass of nitrogen ([tex]\( N_2 \)[/tex]) needed to completely react with [tex]\( 5.8 \)[/tex] grams of hydrogen ([tex]\( H_2 \)[/tex]) according to the balanced chemical equation:
[tex]\[
N_2 + 3H_2 \rightarrow 2NH_3
\][/tex]
Step 1: Determine the moles of hydrogen ([tex]\( H_2 \)[/tex])
First, we need the molar mass of hydrogen gas ([tex]\( H_2 \)[/tex]). The molar mass of a single hydrogen atom (H) is approximately [tex]\( 1 \text{ g/mol} \)[/tex]. Therefore, the molar mass of [tex]\( H_2 \)[/tex] is:
[tex]\[
2 \text{ g/mol}
\][/tex]
Using the mass of [tex]\( H_2 \)[/tex]:
[tex]\[
\text{Mass of } H_2 = 5.8 \text{ g}
\][/tex]
We can find the moles of [tex]\( H_2 \)[/tex] as follows:
[tex]\[
\text{Moles of } H_2 = \frac{\text{Mass of } H_2}{\text{Molar mass of } H_2} = \frac{5.8 \text{ g}}{2 \text{ g/mol}} = 2.9 \text{ moles}
\][/tex]
Step 2: Determine the moles of nitrogen ([tex]\( N_2 \)[/tex]) needed
According to the balanced chemical equation, [tex]\( 1 \)[/tex] mole of [tex]\( N_2 \)[/tex] reacts with [tex]\( 3 \)[/tex] moles of [tex]\( H_2 \)[/tex]. Therefore, the moles of [tex]\( N_2 \)[/tex] needed can be calculated by dividing the moles of [tex]\( H_2 \)[/tex] by [tex]\( 3 \)[/tex]:
[tex]\[
\text{Moles of } N_2 = \frac{\text{Moles of } H_2}{3} = \frac{2.9 \text{ moles}}{3} = 0.9666666666666667 \text{ moles}
\][/tex]
Step 3: Determine the mass of nitrogen ([tex]\( N_2 \)[/tex]) needed
Finally, we need the molar mass of nitrogen gas ([tex]\( N_2 \)[/tex]). The molar mass of a single nitrogen atom (N) is approximately [tex]\( 14 \text{ g/mol} \)[/tex]. Therefore, the molar mass of [tex]\( N_2 \)[/tex] is:
[tex]\[
28 \text{ g/mol}
\][/tex]
Using the moles of [tex]\( N_2 \)[/tex]:
[tex]\[
\text{Moles of } N_2 = 0.9666666666666667 \text{ moles}
\][/tex]
We can find the mass of [tex]\( N_2 \)[/tex] as follows:
[tex]\[
\text{Mass of } N_2 = \text{Moles of } N_2 \times \text{Molar mass of } N_2 = 0.9666666666666667 \text{ moles} \times 28 \text{ g/mol} = 27.066666666666666 \text{ g}
\][/tex]
Therefore, the mass of nitrogen needed to react completely with [tex]\( 5.8 \)[/tex] grams of hydrogen is [tex]\( 27.07 \)[/tex] grams (rounded to two decimal places).
