Respuesta :
Answer: The volume of hydrogen gas formed is 9.5 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 = 7.01 g
Molar mass of aluminium = 27 g/mol
Putting values in above equation, we get:
[tex]\text{Moles of aluminium}=\frac{7.01g}{27g/mol}=0.26mol[/tex]
The given chemical equation follow:
[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 known as excess reagent.
So, aluminium is considered as a 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, 0.26 moles of aluminium will produce = [tex]\frac{3}[2}\times 0.26=0.39mol[/tex] of hydrogen gas
- The equation given by ideal gas follows:
[tex]PV=nRT[/tex]
where,
P = pressure of hydrogen gas = 1.00 atm
V = Volume of hydrogen gas = ? L
T = Temperature of hydrogen gas = [tex]23^oC=[23+273]K=296K[/tex]
R = Gas constant = [tex]0.0821\text{ L. atm }mol^{-1}K^{-1}[/tex]
n = number of moles of hydrogen gas = 0.39 moles
Putting values in above equation, we get:
[tex]1.00atm\times V=0.39mol\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 273K\\\\V=\frac{0.39\times 0.0821\times 296}{1.00}=9.5L[/tex]
Hence, the volume of hydrogen gas formed is 9.5 L
