Question

Consider 2 Al + 6 HCl → 2 AlCl3 + 3 H2 , the reaction of Al with HCl to produce hydrogen gas. What is the pressure of H2 if the hydrogen gas collected occupies 14.0 L at 300.K and was produced upon reaction of 4.50 moles of Al and excess HCl in a process that has a 75.4 percent yield?

Answer 1

Answer : The pressure of hydrogen gas is 8.96 atm.

Explanation :

The given balanced chemical reaction is:

$2Al+6HCl\rightarrow 2AlCl_3+3H_2$

From the balanced chemical reaction we conclude that,

As, 2 moles of Al react to give 3 moles of $H_2$ gas

So, 4.50 moles of Al react to give $\frac{3}{2}\times 4.50=6.75$ moles of $H_2$ gas

Now we have to calculate the moles of $H_2$ gas when percent yield is 75.4.

$\text{The moles of }H_2=75.4\% \times 6.75=\frac{75.4}{100}\times 6.75=5.09moles$

Now we have to calculate the pressure of $H_2$ gas.

Using ideal gas equation :

PV = nRT

where,

P = Pressure of hydrogen gas = ?

V = Volume of the hydrogen gas = 14.0 L

n = number of moles of hydrogen gas = 5.09 moles

R = Gas constant = $0.0821L.atm/mol.K$

T = Temperature of hydrogen gas = 300 K

Putting values in above equation, we get:

$P\times 14.0L=5.09mol\times 0.0821L.atm/mol.K\times 300K\\\\P=8.96atm$

Therefore, the pressure of hydrogen gas is 8.96 atm.