Ammonia is produced by the following reaction. 3h2(g) + n2(g) 2nh3(g) when 7.00 g of hydrogen react with 70.0 g of nitrogen, hydrogen is considered the limiting reactant because

The hydrogen produces the smaller amount of ammonia.  

Step-by-step explanation:

We are given the masses of two reactants, so this is a limiting reactant problem.

We know we will need a balanced equation with masses and molar masses, so let’s gather all the information in one place.  

M_r:      28.02   2.016     17.03

                 N₂  +  3H₂ ⟶ 2NH₃

Mass/g:   70.0    7.00

1. Calculate the moles of N₂ and H₂

Moles N₂ = 70.0 × 1/28.02  

Moles N₂ = 2.498 mol N₂  

Moles H₂ = 7.00 × 2.016  

Moles H₂ = 3.472 mol N₂  

2. Calculate the moles of NH₃ from each reactant

From N₂:

The molar ratio is 2 mol NH₃/1 mol N₂

Moles of NH₃ = 2.498 × 2/1

Moles of NH₃ = 4.996 mol NH₃

From H₂:

The molar ratio is 2 mol NH₃/3 mol H₂  

Moles of NH₃ = 3.472 × 2/3  

Moles of NH₃ = 4.139 mol NH₃

3. Identify the limiting reactant

The limiting reactant is H₂, because it produces fewer moles of NH₃.

its mg hope this

it means there are 3 atoms of oxygen