Question

please answer me I have asked this question two times before please answer me

Answer 1

2.42 gm of copper gave 3.025 gm ofa black oxide of copper 6.49 gm of black oxide on reduction with hydrogen gave 5.192 gm 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

so formula of black oxide is CuO
now,
CuO + H2 ------------------> Cu + H2O

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

moles of cu = 6.49/79.5
ie 6.49 × 63.5/79.5

= 5.192

now you can the prove that the reaction is accordance with law of constant proportion...


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