Question

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Answer 1

⏩⏩⏩2.42g of copper gave 3.025 of a black oxide of copper 6.49 of a black oxide, on reduction with hydrogen, gave 5.192 of copper. Show that these figures are in accordance with law of constant proportion??

Understand the reaction first,

xCuO + ½ O2 ----------------->  Cu(x)O

 x moles of Cu gives 1 mole of Cu(x)O

so 

2.42/63.5 must give 1/x X 2.42/63.5

therefore,

1/x X 2.42/63.5 X (63.5 + 16 X x)

=> 3.025

by solving the above equation we get x = 1

so formula of black oxide is CuO

now,

CuO + H2 ------------------> Cu + H2O

now,

we have 6.49gm of CuO moles = 6.49/79.5 mole

so 

moles of Cu = 6.49/79.5

ie 6.49 X 63.5/79.5

=>5.192