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.
6.06 g
14.01 g
34.08 g
68.16 g
28.02 g

34.08 g

Explanation:

Take the atomic mass of N=14.0, H=1.0/

no. of moles=mmass/molar mass

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

=1.0007mol

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

so one mole of N2 produces 2 moles of NH3.

Using this ratio, we can deduce that the no. of moles of NH3 = 1.0007x2

= 2.001428mol

mass = no. of moles x molar mass

Hence,

mass of NH3 produced = 2.001428 x (14+1x3)

= 34.03

≈C

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