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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. |
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Answer» Solution :Out of Al, Fe, and Cu, Al reacts with NaOH. (a). `2Al+2NaOH+10H_2Oto2Na[Al(OH)_4.H_2O]` `1 mol` of `Al -= 1` " mol of "`NaOH` `-=((13.33)/(40))` " mol of "`NaOH` `-=((13.33)/(40))` " mol of "`Al` `(13.33)/(40)xx27g of Al=9g of Al` (b). `2Al+6"HCl"to2AlCl_3` Cu does not REACT with HCl. 2 " mol of "`Al-=6 mol` of `AlCl_3` `(9)/(27) mol =[(9)/(27)xx3]xx36.5-=36.5-=36.6 g` of HCl is used for Al (c). `Fe+2"HCl"toFeCl_2+H_2` (d). TOTAL moles of HCl required to react with Al and Fe `=MxxV_L` Weight of `HCl=((10xx10xx1.1)/(36.5)xx(343.64)/(1000))xx36.5` =Weight of HCl used for `Fe=37.8-36.5-=1.3g HCl` 2 " mol of "`HCl-=1` " mol of "`Fe=56 of Fe` `((1.3)/(36.5))mol` of `HCl-=((56)/(2)xx(1.3)/(36.5))-=0.997g Fe` All these metals reacts with with `Cl_2` `2Fe+3Cl_2to2FeCl_3` Volume of `Cl_2` at `STP=((3)/(2)xx(0.997)/(56))xx22.4L` `=0.598 L` at `STP` (V). `2Al+3Cl_2to2AlCl_3` Volume of `Cl_2 at STP=((3)/(2)xx(9)/(27))xx22.4` `=11.2 L` of `Cl_2` (vi). `Cu+Cl_2toCuCl_2` Volume of `Cl_2 at STP=(12.5-11.2-0.598)` `=0.702 L of Cl_2` Weight of `Cu=(0.7)/(22.4)xx63.5=1.99 g Cu` `Al=9g,Fe=0.997,Cu=1.99g` Total weight `=11.987g` `%Al=75.09,%Fe=8.31,%Cu=16.6` |
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