Consider preparing ammonia gas by the following reaction: N2(g) + 3H2(g) ===> 2NH3(g)
If you have 28.02 g of N2 and excess H2, how many grams of NH3(g) can be produced.
6.06g
14.01g
34.08 g
68.16 g
28.02g

34.08g

Explanation:

(Take the atomic mass of N=14.0, H=1.0)

no. of moles = mass / molar mass

no. of moles of N2 used = 28.02 / (14x2)

= 1 mol

Since H2 is in excess and N2 is the limiting reagent, the no. of moles of NH3 produced depends on the no. of moles of N2 used.

From the equation, the mole ratio of N2:NH3 = 1: 2,

which means one mole of N2 produces 2 moles of NH3.

From the ratio, we can deduce the no. of moles of NH3 produced = 1 x 2 =2moles.

Mass = no. of moles x molar mass

Mass of NH3 produced = 2 x (14.0+1.0x3)

= 34g

≈34.08g (this happens because the atomic mass I took is only corrected to 1 decimal place, but in the end, this is only a very small error in the decimals. Make sure you take a look if the question provides the atomic mass:) )

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