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6.54 g of zinc are treated with 11.5 g of H_2SO_4. Calculate the volume of hydrogen evolved at S.T.P. How much H_2SO_4 will be left in excess?

Answer» <html><body><p></p>Solution :The <a href="https://interviewquestions.tuteehub.com/tag/corresponding-935567" style="font-weight:bold;" target="_blank" title="Click to know more about CORRESPONDING">CORRESPONDING</a> <a href="https://interviewquestions.tuteehub.com/tag/chemical-914796" style="font-weight:bold;" target="_blank" title="Click to know more about CHEMICAL">CHEMICAL</a> equation is: <br/> `underset(65.38 g)(Zn) + underset(2.016 + 32.06 + 64.0= 98.1 g)(H_(2)SO_(4)) to ZnSO_(4) + underset(22.4 "L at S.T.P")(H_(2))` <br/> From the equation, it is clear that 6.54 g of zinc will react with 9.81 g of `H_2SO_4`. Thus, the amount of `H_2SO_4` <a href="https://interviewquestions.tuteehub.com/tag/taken-659096" style="font-weight:bold;" target="_blank" title="Click to know more about TAKEN">TAKEN</a> is in excess. Hence, Zn will get consumed completely and this is the limiting reagent. <br/> `therefore 65.38 g` of zinc <a href="https://interviewquestions.tuteehub.com/tag/evolve-977729" style="font-weight:bold;" target="_blank" title="Click to know more about EVOLVE">EVOLVE</a> `H_(2) = 22.4 L` at S.T.P. <br/> `therefore 6.54 g` of zinc will evolve <br/> `H_(2) = 22.4/(65.38) xx 6.54 = 2.24` L at S.T.P. <br/> Further, <br/> `therefore 65.38` g of zinc react with `=H_(2)SO_(4) = 98.1 g` <br/> `therefore H_(2)SO_(4)` left in excess `=11.5- 98.1 = 1.69 g` <br/> Hence, 2.24 L hydrogen will be evolved at S.T.R and 1.69 g of `H_2SO_4` will be left in excess.</body></html>


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