An impure sample of zinc (zn) is treated with an excess of sulfuric acid (h 2 so 4) to form zinc sulfate (znso4) and molecular hydrogen (h 2). (a) write a balanced equation for the reaction. (b) if 0.0764 g of h 2 is obtained from 3.86 g of the sample, calculate the percent purity of the sample. (c) what assumptions must you make in (b)

a) The complete balanced chemical equation is:

Zn(s) + H2SO4(aq) > ZnSO4(aq) + H2 (g) 

b) First we find the amount zinc that has reacted based on the given H2 produced.

From stoichiometry 1 moles of Zn is needed for every 1 moles of H2 produced, therefore:

moles(Zn) = moles(H2) 

where moles is the ratio of mass and molar mass (MM)
mass(Zn) / MM(Zn) = mass(H2) / MM(H2) 
mass(Zn) = [mass(H2) / MM(H2)] * MM(Zn) 
mass(Zn) = [(0.0764 g)/(2 g/mol)] * 65.38 g/mol 
mass(Zn) = 2.49 g 

So we got 2.49 g pure zinc in the sample so the purity of zinc is therefore: 

purity = (2.49 / 3.86) * 100 % = 64.50 % 
 

c) In part (b) what we need to assume is that in the impurities in the sample do not react with sulfuric acid to produce hydrogen. So that the hydrogen is only from the reaction of Zn and sulphuric acid.

as given the equation is

2al + 6 hcl > 2 alcl3 + 3 h2(g)

as per equation two moles of aluminum reacts with six moles of hcl to give two moles of alcl3 and three moles of hydrogen gas

given: three moles of al are present and 7 moles of hcl

for three moles of aluminum nine moles of hcl will be required

however as 7 moles of hcl are present so the limiting reagent is hcl

hcl is going to decide the number of moles of product formed

when six moles of hcl are present = 3 moles of h2 are formed

when one mole of hcl is present == 3/6 mole sof h2 will be formed

when 7 moles of hcl are present = 3 x 7/ 6 moles of h2 will be formed = 3.5 moles of h2