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10 mL of H_(2) combine with 5 mL of O_(2) to form H_(2)O when 200 mL of H_(2) at S.T.P. is passed through heated CuO, latter loses 0.144 g of its weight. Does the above data correspond to the law of constant composition ?

Answer» <html><body><p><br/></p>Solution :In the first experiment : <br/> Ratio by volume of `H_(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>) and O_(2)` <a href="https://interviewquestions.tuteehub.com/tag/combining-922870" style="font-weight:bold;" target="_blank" title="Click to know more about COMBINING">COMBINING</a> to form `H_(2)O=2 :1` <br/> In the second experiment : <br/> `CuO+H_(2)overset(heat)(rarr)Cu+H_(2)O` <br/> The loss weight of CuO is due to oxygen which has been removed. It is 0.44 g <br/> Now, 32 g of `O_(2)` at S.T.P occupy = 22400 L <br/> 0.144 g of `O_(2)` at S.T.P occupy `= (("22400 mL"))/(("32 g"))xx(0.144g)=100.8mL` <br/> Ratio by volume of `H_(2) and O_(2)` combining to form `H_(2)O = <a href="https://interviewquestions.tuteehub.com/tag/200-288914" style="font-weight:bold;" target="_blank" title="Click to know more about 200">200</a> : 100.8 = 2:1` <br/> Since the <a href="https://interviewquestions.tuteehub.com/tag/two-714195" style="font-weight:bold;" target="_blank" title="Click to know more about TWO">TWO</a> ratios are the same, the data <a href="https://interviewquestions.tuteehub.com/tag/corresponds-935737" style="font-weight:bold;" target="_blank" title="Click to know more about CORRESPONDS">CORRESPONDS</a> to Law of Constant Composition.</body></html>


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