15. What mass of nitrogen is needed to react completely with 5.8 g of hydrogen according to this balanced chemical equation?

[tex]\[ N_2 + 3 H_2 \rightarrow 2 NH_3 \][/tex]

A. 81 g [tex]\(N_2\)[/tex]

B. 76 g [tex]\(N_2\)[/tex]

C. 27 g [tex]\(N_2\)[/tex]

D. 5 g [tex]\(NH_3\)[/tex]



Answer :

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).