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The balance chemical reaction is 2Al + 6HCl -à 2AlCl3 + 3H2 Assuming excess amount of HCl because it is not given Mole Al reacted = 13.5 g ( 1 mole/ 27 g) = 0.5 mole Al
Mole H2 = 0.5 mol Al ( 3 mole H2 / 2 mole Al) = 0.75 mole H2
since at STP then 1 mole gas occupies 22.414 L volume H2 = 0.75 mole (22.414 L / mol) = 16.8105 L H2
Have a nice day
The balance chemical reaction is 2Al + 6HCl -à 2AlCl3 + 3H2 Assuming excess amount of HCl because it is not given Mole Al reacted = 13.5 g ( 1 mole/ 27 g) = 0.5 mole Al
Mole H2 = 0.5 mol Al ( 3 mole H2 / 2 mole Al) = 0.75 mole H2
since at STP then 1 mole gas occupies 22.414 L volume H2 = 0.75 mole (22.414 L / mol) = 16.8105 L H2
Have a nice day
Answer : The moles of Al is, 0.50 mole.
The moles of [tex]H_2[/tex] is, 0.75 mole.
The volume of [tex]H_2[/tex] is, 16.8 liters.
Explanation : Given,
Mass of aluminum = 13.5 g
Molar mass of aluminum = 26.98 g/mole
First we have to calculate the moles of aluminum.
[tex]\text{Moles of }Al=\frac{\text{Mass of }Al}{\text{Molar mass of }Al}=\frac{13.5g}{26.98g/mole}=0.50moles[/tex]
The moles of Al is, 0.50 mole.
Now we have to calculate the moles of [tex]H_2[/tex].
The balanced chemical reaction will be,
[tex]2Al(s)+6HCl(l)\rightarrow 2AlCl_3(s)+3H_2(g)[/tex]
From the balanced reaction, we conclude that
As, 2 moles of Al react to give 3 moles of [tex]H_2[/tex]
So, 0.50 moles of Al react to give [tex]\frac{3}{2}\times 0.50=0.75[/tex] moles of [tex]H_2[/tex]
The moles of [tex]H_2[/tex] is, 0.75 mole.
Now we have to calculate the volume of [tex]H_2[/tex].
At STP,
As, 1 mole of [tex]H_2[/tex] contains 22.4 L volume of [tex]H_2[/tex]
So, 0.75 mole of [tex]H_2[/tex] contains [tex]0.75\times 22.4=16.8L[/tex] volume of [tex]H_2[/tex]
The volume of [tex]H_2[/tex] is, 16.8 liters.
