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InterviewSolution
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When 1.82 g a of mixture of Al and an unkown metal. Arranged in the series of the standard reduction electrode potential below hydrogen, was dissolved in `HCl`, 0.672 L of `H_2`, measured at STP was liberated. To oxidise this mixture, 0.56 L of `O_2` measured at STP was needed which unknown metal was taken? Determine the mass percentage of metals in the mixture. |
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Answer» (a). Only Al gives `H_2`, not the other metal below `H_2`. `2Al+6"HCl"to2AlCl_3+3H_2` Ew of `H_2=" Eq of "Al (xg)` `(0.672)/(11.2)=((x)/(27))/(3)` `(1 " mol of "H_2=2 " Eq of "H_2=2.4L)` `(1 " Eq of "H_2=11.2L)` `thereforex=0.54 of Al` The weight of unknown metal `(M)=1.82-0.54=1.28g` (b). `4Al+3O_2to2Al_2O_3` `{:(4 mol Al, 3 " mol of "O_2),(4xx27g,3xx32g O_2):}` `4xx27g of Al implies3xx32g O_2` `0.54 of "Al "implies(3xx32)/(4xx27)xx0.54 g of O_2` `(1 " mol of "O_2=32g=22.4L)` Volume of `O_2=(22.4)/(32)xx(3)/(4)xx(32)/(27)xx0.54` `=0.336 L of O_2` `O_2` used for metal `M=(0.56-0.336)L=0.224L` Let M be monovalent. `1 " mol of "M= 1 " mol of "O_2` `implies(1.28)/("Atomic weight")=(0.224)/(22.4)" mol of "O_2` `therefore` Atomic weight`=(1.28)/(0.224)xx22.4` `=128` which is not possible So the metal is divelent `2 mol=1 " mol of "O_2` `therefore` Atomic weight `=(128)/(2)=64` i.e., metal is copper. |
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