Respuesta :
Answer: The volume of hydrogen gas produced is 4.24 L
Explanation:
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
Given mass of aluminium = 3.12 g
Molar mass of aluminium = 26.98 g/mol
Putting values in above equation, we get:
[tex]\text{Moles of aluminium}=\frac{3.12g}{26.98g/mol}=0.116mol[/tex]
For the given chemical reaction:
[tex]2NaOH(aq.)+2Al(s)+6H_2O(l)\rightarrow 2NaAl(OH)_4(aq.)+3H_2(g)[/tex]
As, NaOH is present in excess. It is considered as an excess reagent.
Aluminium is the limiting reagent because it limits the formation of product.
By Stoichiometry of the reaction:
2 moles of aluminium produces 3 moles of hydrogen gas
So, 0.116 moles of aluminium will produce = [tex]\frac{3}{2}\times 0.116=0.174mol[/tex] of hydrogen gas
To calculate the volume of hydrogen gas produced, we use the ideal gas equation:
[tex]PV=nRT[/tex]
where,
P = pressure of the gas = 750 torr
V = Volume of the gas = ?
T = Temperature of the gas = [tex]20^oC=[20+273]K=293K[/tex]
R = Gas constant = [tex]62.364\text{ L. Torr }mol^{-1}K^{-1}[/tex]
n = number of moles of the gas = 0.174 mol
Putting values in above equation, we get:
[tex]750torr\times V=0.174mol\times 62.364\text{ L Torr }mol^{-1}K^{-1}\times 293K\\n_{mix}=\frac{0.174mol\times 62.364\text{ L Torr }mol^{-1}K^{-1}\times 293K}{750torr}=4.24L[/tex]
Hence, the volume of hydrogen gas produced is 4.24 L
