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Q) 1 g mixture of cuprous oxide and cupric oxide was quantitatively reduced to 0.839 g of metallic copper. What was the weight of cupric oxide in the original sample? (Cu = 63.5, O = 16)
Asked by Anish | 07 Oct, 2018, 12:20: AM
answered-by-expert Expert Answer
The reaction equations are:
 
Atomic space weight space of space Cu space equals space 63.5 Molecular space weight space of space Cu subscript 2 straight O space equals space 143 Molecular space weight space of space CuO space equals 79.5 Let comma Moles space of space Cu subscript 2 straight O space equals space straight x space Moles space of space CuO space equals space straight y therefore 143 straight x space plus space 79.5 straight y space equals 1 and 127 straight x space plus space 63.5 space straight y space equals space 0.839 after space solving semicolon straight y space equals space 0.0069 So comma space weight space of space CuO space equals space 79.5 cross times space 0.0069 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals 0.548 space straight g Weight space of space space Cu subscript 2 straight O space equals space 0.45 space straight g
The weight of cupric oxide in the sample is 0.548 g

 
 
Answered by Varsha | 10 Oct, 2018, 09:45: AM

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