`Al+3H^+ -gt Al^(3+)+ 3/2H_2` How many grams of `H_2`  is formed by dissolving 900mg of Al in dilute HCl.  



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Posted on (Answer #1)

`Al+3H^+ rarr Al^(3+)+3/2H_2`


Mole ratio:

`Al:H_2 = 1:3/2`


molar mass of Aluminium = 27g/mol

Amount of Al added `= 0.9/27 mol = 1/30mol`

Amount of `H_2` gas produced `= 3/2xx1/30 = 1/20mol`


Molar mass of `H_2` gas = 2g/mol

Mass of `H_2 ` formed `= 2xx1/20 = 1/10g = 0.1g`


So the `H_2` gas formed are 0.1g.



There is excess HCl for the complete reaction


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