Question

Consider the reaction.

N2(g)+3H2(g) ---> 2NH3(g)
How many grams of N2 are required to produce 100.0 L of NH3 at STP?

A.6.25 g
B.12.5 g
C.62.5 g
D.125.0 g

Answer 1

C.62.5 g

Step-by-step explanation:

At STP,

Pressure = 1 atm

Temperature = 273.15 K

Given, Volume = 100.0 L

Using ideal gas equation as:

`PV=nRT`

where,

P is the pressure

V is the volume

n is the number of moles

T is the temperature

R is Gas constant having value = 0.0821 L atm/ K mol

Applying the equation as:

1 atm × 100.0 L = n ×0.0821 L atm/ K mol × 273.15 K

⇒n = 4.4592 moles

From the given reaction,

`N_2_((g))+3H_2_((g))\rightarrow 2NH_3_((g))`

2 moles of ammonia is formed when 1 mole of nitrogen gas undergoes reaction.

Also, 1 mole of ammonia is formed when 1/2 mole of nitrogen gas undergoes reaction.

4.0446 moles of ammonia is formed when 1/2*4.4592 moles of nitrogen gas undergoes reaction.

Moles of nitrogen = 2.2296 moles

Molar mass of nitrogen,
`N_2` = 28.0134 g/mol

The formula for the calculation of moles is shown below:

`moles = (Mass\ taken)/(Molar\ mass)`

Thus,

`2.2296\ moles= (Mass)/(28.0134\ g/mol)`

Mass = 62.5 g

Answer 2

Answer:The correct answer is option C.

Step-by-step explanation:

`N_2(g)+3H_2(g)\rightarrow 2NH_3(g)`

At STP, 1 mol of gas occupies 22.4 L of volume

So, 100.0 L volume of
`NH_3`will be occupied by :

`(1)/(22.4 L)* 100.0 L=4.4642 moles`

According to reaction 2 moles
`NH_3` are obtained from 1 mole of
`N_2`.

Then,4.4642 moles of
`NH_3` will be obtained from :

`(1)/(2)* 4.4642` that is 2.2321 moles of
`N_2`

Mass of
`N_2` gas:

Moles of
`N_2` gas Molar mass of
`N_2` gas:

`2.2321 mol* 28 g/mol=62.4988\approx 62.50 g`

Hence, the correct answer is option C.