Problem: 3 H2 (g) + N2 (g)→ 2 NH3(g)a. How many moles of NH3 can be produced from 13.5 mol of H2 and excess N2 ?b. How many grams of NH3 can be produced from 4.90 mol of N2 and excess H2 .c. How many grams of H2 are needed to produce 12.74 g of NH3 ?d. How many molecules (not moles) of NH3 are produced from 1.05×10−4 g of H2 ?

FREE Expert Solution

Balanced Reaction: 

3 H(g) + N(g)→ 2 NH3(g)


a. Moles of NH3 from 13.5 mol of Hand excess N

Since N2 is excess, calculate directly from H2.


13.5 mol H2 ×2 mol NH23 mol H2  

= 9 mol NH

9 moles of NH3 are produced from 3.5 mol of Hand excess N



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Problem Details

3 H(g) + N(g)→ 2 NH3(g)

a. How many moles of NHcan be produced from 13.5 mol of Hand excess N?

b. How many grams of NHcan be produced from 4.90 mol of Nand excess H.

c. How many grams of Hare needed to produce 12.74 g of NH?

d. How many molecules (not moles) of NHare produced from 1.05×10−4 g of H?

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