In order to solve the question b):
We have the following reaction occuring:
[tex]2Al+6HCl\rightarrow2AlCl_3+3H_2[/tex]
We need to calculate the number of moles of aluminum needed to produce 30ml of H2 (hydrogen gas) at a temperature of 22°C and a pressure of 763mmHg.
To calculate the number of moles of hydrogen produced we use the ideal gas equation:
[tex]P.V=n.R.T[/tex]
Where:
P is the pressure
V is the volume
n is the number of moles
R is the gas constant
T is the temperature
R is a constant and it's value is:
[tex]R=8.314\text{ }\frac{m^3.Pa}{K.mol}[/tex]
So we need to convert each variable to the units of this constant.
[tex]\begin{gathered} T:\text{ }22^{\circ}C=295.15^{\circ}K \\ P:\text{ }763mmHg=101725\text{ }Pa \\ V:\text{ 30ml}:0.00003m^3 \end{gathered}[/tex]
So now we calculate:
[tex]n=\frac{P.V}{R.T}=\frac{101725Pa.0.00003m^3}{8.314\frac{m^3.Pa}{K.mol}.295.15^{\circ}C}=0.00124mol[/tex]
So we know that 0.00124 moles of H2 are formed.
Now we know that for every 2 moles of aluminum 3 moles of H2 are formed.
So we calculate the moles of Al needed:
[tex]n_{Al}=0.00124mol_{H2}.\frac{2mol_{Al}}{3mol_{H2}}=0.000826mol_{Al}[/tex]
So the answer is 0.000826 moles of Al°
