Question

At 600 K and 1.75 atm, if 16.00 grams of H2 will reacts with excess N2 what volume in liters will be produced of NH3?

N2(g) + 3 H2(g) --> 2 NH3(g)
(R = 0.0821 L atm/mol K)

Answer 1

Answer : The volume of
`NH_3` produced will be, 150.0 L

Explanation :

First we have to calculate the moles of
`H_2`

`\text{Moles of }H_2=\frac{\text{Given mass }H_2}{\text{Molar mass }H_2}=(16.00g)/(2g/mol)=8mol`

Now we have to calculate the moles of
`NH_3`

The balanced chemical equation is:

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

From the reaction, we conclude that

As, 3 moles of
`H_2` react to give 2 moles of
`NH_3`

So, 8 moles of
`H_2` react to give
`(2)/(3)* 8=5.33` mole of
`NH_3`

Now we have to calculate the volume of
`NH_3`

Using ideal gas equation:

PV = nRT

where,

P = pressure of gas = 1.75 atm

V = volume of gas = ?

n = number of moles of gas = 5.33 mol

T = temperature of gas = 600 K

R = gas constant = 0.0821 L.atm/mol.K

Now put all the given value in the above formula, we get:

`1.75atm* V=5.33mol* 0.0821 L.atm/mol.K* 600K`

`V=150.0L`

Therefore, the volume of
`NH_3` produced will be, 150.0 L