This is an incomplete question, here is a complete question.
Suppose we now collect hydrogen gas, H₂(g), over water at 21°C in a vessel with total pressure of 743 Torr. If the hydrogen gas is produced by the reaction of aluminum with hydrochloric acid:
![2Al(s)+6HCl(aq)\rightarrow 2AlCl_3(aq)+3H_2(g)](https://tex.z-dn.net/?f=2Al%28s%29%2B6HCl%28aq%29%5Crightarrow%202AlCl_3%28aq%29%2B3H_2%28g%29)
what volume of hydrogen gas will be collected if 1.35 g Al(s) reacts with excess HCl(aq)? Express your answer in liters.
Answer : The volume of hydrogen gas that will be collected is 1.85 L
Explanation :
First we have to calculate the number of moles of aluminium.
Given mass of aluminium = 1.35 g
Molar mass of aluminium = 27 g/mol
![\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}](https://tex.z-dn.net/?f=%5Ctext%7BNumber%20of%20moles%7D%3D%5Cfrac%7B%5Ctext%7BGiven%20mass%7D%7D%7B%5Ctext%7BMolar%20mass%7D%7D)
![\text{Moles of aluminium}=\frac{1.35g}{27g/mol}=0.05mol](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20aluminium%7D%3D%5Cfrac%7B1.35g%7D%7B27g%2Fmol%7D%3D0.05mol)
The given chemical reaction is:
![2Al(s)+6HCl(aq)\rightarrow 2AlCl_3(aq)+3H_2(g)](https://tex.z-dn.net/?f=2Al%28s%29%2B6HCl%28aq%29%5Crightarrow%202AlCl_3%28aq%29%2B3H_2%28g%29)
As, hydrochloric acid is present in excess. So, it is considered as an excess reagent.
Thus, aluminium is a limiting reagent because it limits the formation of products.
By Stoichiometry of the reaction:
2 moles of aluminium produces 3 moles of hydrogen gas
So, 0.005 moles of aluminium will produce =
of hydrogen gas
Now we have to calculate the mass of helium gas by using ideal gas equation.
PV = nRT
where,
P = Pressure of hydrogen gas = 743 Torr
V = Volume of the helium gas = ?
n = number of moles of hydrogen gas = 0.075 mol
R = Gas constant = ![62.364\text{ L Torr }mol^{-1}K^{-1}](https://tex.z-dn.net/?f=62.364%5Ctext%7B%20L%20Torr%20%7Dmol%5E%7B-1%7DK%5E%7B-1%7D)
T = Temperature of hydrogen gas = ![21^oC=[21+273]K=294K](https://tex.z-dn.net/?f=21%5EoC%3D%5B21%2B273%5DK%3D294K)
Now put all the given values in above equation, we get:
![743Torr\times V=0.075mol\times 62.364\text{ L Torr }mol^{-1}K^{-1}\times 294K\\\\V=1.85L](https://tex.z-dn.net/?f=743Torr%5Ctimes%20V%3D0.075mol%5Ctimes%2062.364%5Ctext%7B%20L%20Torr%20%7Dmol%5E%7B-1%7DK%5E%7B-1%7D%5Ctimes%20294K%5C%5C%5C%5CV%3D1.85L)
Hence, the volume of hydrogen gas that will be collected is 1.85 L