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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?

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?

Grade:9

1 Answers

varun
304 Points
6 years ago
HELLO NEERAJ,
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
so by these figures you can prove that the reaction is accordance with law of constant proportions
THANK YOU

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