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 $H_2$

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

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 $\frac{2}{3}\times 8=5.33$ 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:

$1.75atm\times V=5.33mol\times 0.0821 L.atm/mol.K\times 600K$

$V=150.0L$

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