Question

How many moles of NH3 from when 32.4 L of H2 gas completely reacts at STP according to the following reaction? Remember 1 mol of an ideal gas has a volume of 22.4 L at STP
N₂(g)+3H₂(g)→2NH₃(g)

ps: We have to not only give the answer which is 1.45 mol NH3 but we have to show the equation of how we got to it.

Answer 1

N_2 (g) + 3H_2 (g) rightarrow 2NH_3 (g) volume of H_2 = 32.44 At STP 1 mole of H_2 = 22.4L ? mole of H_2 = 32.4L therefore moles of H_2

Answer 2

Following are the steps to calculate the volume:

The volume of $\bold{H_2=32.40 \ L}$ at STP is used to get the mole number.

1 mole of $\bold{H_2}$ takes up 22.4 L of $\bold{H_2}$ volume.

1 mole of $\bold{H_2}$ occupies a volume of 32.40 L of $\bold{H_2}$ .

$\bold{H_2}$  mole $\bold{= (\frac{1}{22.4}) \times 32.40 }$

                 $\bold{= 1.45}$

balancing the chemical equation:

$\bold{1\ \ N_2 + 3 H_2 \longrightarrow 2 NH_3}$

The number of moles is calculated as follows:

$\bold{ H_2 \ Mole = 1.45 \ mol}$

Equation of balance:

3 moles of $\bold{H_2}$ produces 2 moles of $\bold{NH_3 }$.

1.45 moles of $\bold{H_2}$ produce ? moles of $\bold{NH_3 }$.

$\bold{NH_3 }$ moles $\bold{= (\frac{2}{3} )\times 1.45}$

                      $\bold{= 0.96\ mol}$

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