Ammonia can also be synthesized by the reaction 3h2(g)+n2(g)→2nh3(g) what is the theoretical yield of ammonia, in kilograms, that we can synthesize from 5.22 kg of h2 and 31.5 kg of n2? express the mass in kilograms to three significant figures.

We are given the complete reaction:

3 H2(g)  +  N2(g) → 2 NH3(g)

First let us convert mass to moles.

moles H2 = 5.22 kg / (2 kg/kmol) = 2.61 kmol H2

moles N2 = 31.5 kg / (28 kg/kmol) = 1.125 kmol N2

Then we find for the limiting reactant. The limiting reactant is the one who has lower (moles/coefficient) ratio.

H2 = 2.61 / 3 = 0.87

N2 = 1.125 / 1 = 1.125

Hence the H2 is the limiting reactant so we should base the calculation of NH3 from it. We see that 2 moles of NH3 is produced for every 3 moles of H2, therefore:

moles NH3 = 2.61 kmol H2 * (2 kmol NH3 / 3 kmol H2) = 1.74 kmol

The molar mass of NH3 is 17 kg/kmol, therefore the mass NH3 is:

mass NH3 = 1.74 kmol * 17 kg/kmol

mass NH3 = 29.58 kg

idk i’m struggling too