Firstly,write the equation between zinc and hcl
Zn + 2HCl -> ZnCl2 + H2
Then calculate the no. of mol available
no.of mol of Zn available= 0.65/65.4=0.00994( 3 sig fig)
no. of mol of HCl available in 50cm^3= 50/1000 x 2= 0.1
determine the limiting reagent and excess
Mol ratio
1Zn: 2HCl
no. of mol of hcl required if all Zn is used up= 2 x
0.00994=0.01988
Hence,
the no. of mol of hcl required is < no. of mol of hcl
available
thus Hcl is in excess so Zn is the limiting reagent
Now to find the volume of H2 gas produced, find the no. of mol
of H2 first
Mol ratio
Zn:H2
no. of mol of H2=0.00994
assuming its at r.t.p
vol. of H2= 0.00994 x 24=0.239dm3 (3 sig fig)
if its at s.t.p
vol. of h2= 0.00994 x 22.4= 0.223dm3 ( 3 sig fig)
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