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A mixutre of Cu, Fe, and Al was reacted with 13.33 g of NaOH. During chlorination with the same amount of meta mixture entered into reaction with 12.5 L of chlorine measured at STP, while for treating the same amound of the metal mixture at STP, while for treating the same amound of the metal mix mixutre 343.64 " mL of " HCl, havig a density of .1g mL^(-1) and containing 10% by mass of HCl were required. Determine the mass percentage of the metals in the mixture.

Answer» <html><body><p></p>Solution :Out of Al, Fe, and Cu, Al reacts with NaOH. <br/> (a). `2Al+2NaOH+10H_2Oto2Na[Al(OH)_4.H_2O]` <br/> `1 mol` of `Al -= 1` " mol of "`NaOH` <br/> `-=((13.33)/(40))` " mol of "`NaOH` <br/> `-=((13.33)/(40))` " mol of "`Al` <br/> `(13.33)/(40)xx27g of Al=9g of Al` <br/> (b). `2Al+6"HCl"to2AlCl_3` <br/> Cu does not <a href="https://interviewquestions.tuteehub.com/tag/react-613674" style="font-weight:bold;" target="_blank" title="Click to know more about REACT">REACT</a> with HCl. <br/> 2 " mol of "`Al-=6 mol` of `AlCl_3` <br/> `(9)/(<a href="https://interviewquestions.tuteehub.com/tag/27-298706" style="font-weight:bold;" target="_blank" title="Click to know more about 27">27</a>) mol =[(9)/(27)xx3]xx36.5-=36.5-=36.6 g` of HCl is used for Al <br/> (c). `Fe+2"HCl"toFeCl_2+H_2` <br/> (d). <a href="https://interviewquestions.tuteehub.com/tag/total-711110" style="font-weight:bold;" target="_blank" title="Click to know more about TOTAL">TOTAL</a> moles of HCl required to react with Al and Fe <br/> `=MxxV_L` <br/> Weight of `HCl=((10xx10xx1.1)/(36.5)xx(343.64)/(1000))xx36.5` <br/> =Weight of HCl used for `Fe=37.8-36.5-=1.3g HCl` <br/> 2 " mol of "`HCl-=1` " mol of "`Fe=56 of Fe` <br/> `((1.3)/(36.5))mol` of `HCl-=((56)/(2)xx(1.3)/(36.5))-=0.997g Fe` <br/> All these metals reacts with with `Cl_2` <br/> `2Fe+3Cl_2to2FeCl_3` <br/> Volume of `Cl_2` at `<a href="https://interviewquestions.tuteehub.com/tag/stp-633132" style="font-weight:bold;" target="_blank" title="Click to know more about STP">STP</a>=((3)/(2)xx(0.997)/(56))xx22.4L` <br/> `=0.598 L` at `STP` <br/> (<a href="https://interviewquestions.tuteehub.com/tag/v-722631" style="font-weight:bold;" target="_blank" title="Click to know more about V">V</a>). `2Al+3Cl_2to2AlCl_3` <br/> Volume of `Cl_2 at STP=((3)/(2)xx(9)/(27))xx22.4` <br/> `=11.2 L` of `Cl_2` <br/> (vi). `Cu+Cl_2toCuCl_2` <br/> Volume of `Cl_2 at STP=(12.5-11.2-0.598)` <br/> `=0.702 L of Cl_2` <br/> Weight of `Cu=(0.7)/(22.4)xx63.5=1.99 g Cu` <br/> `Al=9g,Fe=0.997,Cu=1.99g` <br/> Total weight `=11.987g` <br/> `%Al=75.09,%Fe=8.31,%Cu=16.6`</body></html>


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