StoichiometryMedium

Limiting reagent

For the reaction

\mathrm{N_2 + 3H_2 \to 2NH_3}

, you have 10.0 g of

\mathrm{N_2}

and 5.00 g of

\mathrm{H_2}

. Find the limiting reagent and the moles of

\mathrm{NH_3}

produced.
M(N₂) = 28.0 g/mol, M(H₂) = 2.02 g/mol.

Given data

m(N₂) = 10.0 gm(H₂) = 5.00 gM(N₂) = 28.0 g/molM(H₂) = 2.02 g/mol

Steps

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Calculate the moles of N₂ available.
Write your working — the numerical result must appear in the text mol
Calculate the moles of H₂ available.
Calculate the moles of NH₃ producible from N₂ (ratio 1:2).

Full worked solution

Calculate the moles of N₂ available.
$n(N_2) = \frac{10.0}{28.0}$
$n(\mathrm{N_2}) = m/M = 10.0/28.0 = \mathbf{0.357\,mol}$ . The moles of N₂ are obtained by dividing the mass by the molar mass.
Calculate the moles of H₂ available.
$n(H_2) = \frac{5.00}{2.02}$
$n(\mathrm{H_2}) = m/M = 5.00/2.02 = \mathbf{2.48\,mol}$ . The moles of H₂ are obtained by dividing the mass by the molar mass.
Calculate the moles of NH₃ producible from N₂ (ratio 1:2).
$n(NH_3) = 2 \times 0.357$
$n(\mathrm{NH_3}) = 2 \times n(\mathrm{N_2}) = 2 \times 0.357 = \mathbf{0.714\,mol}$ . N₂ is the limiting reagent; from the 1:2 stoichiometric ratio, 0.714 mol of NH₃ are produced.

Result:N₂ is the limiting reagent. 0.714 mol of NH₃ are produced.

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