1.

A 1-g mixture of cuprous oxideand cupric oxide was quantitatively reducedto 0.839 g of metallic copper . Whatwas t he weightof cupricoxidein the original sample ? (Cu = 63.5, O = 16)

Answer»

Solution :Letthe WEIGHTOF `CuO be x G`. The weight of `Cu_(2) O` will be (1 - x) g . Asthe Cu atomsareconserve,applyingPOAC for Cu atoms,
molesof Cu in CuO + moles of Cu in `Cu_(2)O`
= moles of Cu in theproduct
`1 xx ` moles of `CuO + 2 xx "moles of " Cu_(2) O` = moles of Cu
`(x)/( 79 . 5) + 2 xx (1 - x)/( 143) = (0.839)/( 63.5)[{:(CuO = 79 . 5),(Cu_(2) O = 143):}] `
x = 0.55 g


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