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?

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varun · Answer

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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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?

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