Problem: a. How many moles of NH3 can be produced from 16.5 mol of H 2 and excess N2?b. How many grams of NH3 can be produced from 3.78 mol of N 2 and excess H2.c. How many grams of H 2 are needed to produce 11.05 g of NH3 ?d. How many molecules (not moles) of NH3 are produced from 2.92 × 10−4 g of H2?

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FREE Expert Solution

The balanced reaction for this will appear as:

N₂ (g) + 3 H₂ (g) → 2 NH₃ (g)

a. moles of NH₃:

$\mathbf{mol}\mathbf{ }{\mathbf{NH}}_{\mathbf{3}}\mathbf{ }\mathbf{=}\mathbf{ }\mathbf{16}\mathbf{.}\mathbf{5}\mathbf{ }\cancel{\mathbf{mol}\mathbf{ }{\mathbf{H}}_{\mathbf{2}}}\mathbf{ }\left(\frac{2 \mathrm{mol} {\mathrm{NH}}_3}{3 \cancel{\mathrm{mol} H_2}}\right)$

mol NH₃ = 11 mol NH₃