How many moles of nh3 can be produced from 12.0 mol of h2 and excess n2? express your answer numerically in moles. view available hint(s) nothing mol nh3m o l n h 3 part b how many grams of nh3 can be produced from 4.04 mol of n2 and excess h2. express your answer numerically in grams. view available hint(s) nothing g nh3g n h 3 part c how many grams of h2 are needed to produce 13.02 g of nh3? express your answer numerically in grams. view available hint(s) nothing g h2g h 2 part d how many molecules (not moles) of nh3 are produced from 7.62×10−4 g of h2? express your answer numerically as the number of molecules. view available hint(s) nothing moleculesm o l e c u l e s

A) 8.00 mol NH₃

B) 137 g NH₃

C) 2.30 g H₂

D) 1.53 x 10²⁰ molecules NH₃

Explanation:

Let us consider the balanced equation:

N₂(g) + 3 H₂(g) ⇄ 2 NH₃(g)

Part A

3 moles of H₂ form 2 moles of NH₃. So, for 12.0 moles of H₂:

Part B:

1 mole of N₂ forms 2 moles of NH₃. And each mole of NH₃ has a mass of 17.0 g (molar mass). So, for 4.04 moles of N₂:

Part C:

According to the balanced equation 6.00 g of H₂ form 34.0 g of NH₃. So, for 13.02g of NH₃:

Part D:

6.00 g of H₂ form 2 moles of NH₃. An each mole of NH₃ has 6.02 x 10²³ molecules of NH₃ (Avogadro number). So, for 7.62×10⁻⁴ g of H₂:

a 20-cm-long, cylindrical graphite (pure carbon) rod is inserted into an oxidizing atmosphere at 1145 k and 1.013 x 105 pa pressure. the oxidizing process is limited by the diffu- sion of oxygen counterflow to the carbon monoxide that is formed on the cylindrical surface.