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0.7 of zinc dust containing Zn and ZnO ...

`0.7 ` of zinc dust containing Zn and ZnO when dissolved in dilute `H_(2)SO_(4)` evolved `224 mL` of `H_(2)` at `N.T.P.` Calculate the percentage of zinc in the dust (Atomic mass of `Zn = 65`).

Text Solution

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Both Zn and ZnO present in zinc dust react with dilute `H_(2)SO_(4)`. But only Zn librates hydrogen gas
`{:(ZnO+H_(2)SO_(4),rarr,ZnSO_(4)+,H_(2)O),(Zn+H_(2)SO_(4),rarr,ZnSO_(4)+,H_(2)),(65g,,,22400 mL):}`
Now, `22400` mL of `H_(2)` are evolved from `Zn = 65g`
`224 mL` of `H_(2)` are evolved from Zn `= (65)/(22400) xx 224 = 0.65 g`
`%` of zinc in zinc dust `= ("Mass of zin"')/("Mass of zinc dust") xx 100 = (0.65 g)/(0.70 g) xx 100 = 92.86 g`
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