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.

c. because step up transformer steps up the input voltage i.e increases the input voltage  

the number of turns are greater on the secondary side of step up transformer  

transformer does not handle dc current