Respuesta :
How many grams of h2 gas can be produced by the reaction of 63.0 grams of al(s) with an excess of dilute hydrochloric acid in the reaction shown below? 2 al(s) + 6 hcl(aq) → 2 alcl3(aq) + 3 h2(g)?
Answer: The mass of hydrogen gas produced by the reaction is 6.9 grams
Explanation:
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] .....(1)
Given mass of aluminium = 63 g
Molar mass of aluminium = 27 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of aluminium}=\frac{63g}{27g/mol}=2.33mol[/tex]
For the given chemical reaction:
[tex]2Al(s)+6HCl(aq.)\rightarrow 2AlCl_3(aq.)+3H_2(g)[/tex]
Hydrochloric acid is present in excess. So, it is considered as an excess reagent. And, aluminium metal is a limiting reagent because it limits the formation of product.
By Stoichiometry of the reaction:
2 moles of aluminium metal produces 3 moles of hydrogen gas.
So, 2.33 moles of aluminium metal will produce = [tex]\frac{3}{2}\times 2.33=3.45mol[/tex] of hydrogen gas
Now, calculating the mass of hydrogen gas by using equation 1:
Moles of hydrogen gas = 3.45 moles
Molar mass of hydrogen gas = 2 g/mol
Putting values in equation 1, we get:
[tex]3.45mol=\frac{\text{Mass of hydrogen gas}}{2g/mol}\\\\\text{Mass of hydrogen gas}=(3.45mol\times 2g/mol)=6.9g[/tex]
Hence, the mass of hydrogen gas produced by the reaction is 6.9 grams
