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.

first let us lay out the formula for potential

and kinetic energy:

pe = mgh

ke = (mv^2)/2

where: m=mass, v=velocity, g=gravitational

acceleration, h=height

calculating for mass using the known values of

pe:

340j = m (9.81 m/s^2) (50ft) (1m/3.28ft)

m = 2.27 kg

the law of conservation of energy states that

energy is neither created nor destroyed. therefore, the change in kinetic

energy from 50 ft to 35 ft would be equal to the change in potential energy

from 50 ft to 35 ft. however it would be opposite in signs since one is losing

while other is gaining.

ke(35ft) – ke(50ft) = - [ pe(35ft) – pe(50ft]

ke(50ft) = 0 since the brick is initially at rest

ke(35ft) = 340j - 2.27kg (9.81 m/s^2) (35ft)

(1m/3.28 ft)

ke(35ft)

= 102.38 j