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One gram of an alloy of Al and Mg when t...

One gram of an alloy of `Al and Mg` when treated with excess of dilute `HCI` forms `MgCI_2, AICI_3` and hydrogen. The evolved hydrogen collected over Hg at 0°C has a volume of `1.20` litres at `0.92` atm pressure. Calculate the composition of the alloy. `("AI" = 27, "Mg" = 24)`

Text Solution

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Volume of `H_2` produced at NTP `= ( 1.2 xx 0.92)/( 273) xx (273)/(1)`
`((p_1 V_1)/( T_1) = ( p_2 V_2)/( T_2) )`
`= 1.104` litres.
Since both `Al and Mg` produce `1.104` litres of `H_2`,
equivalent of `Al+` equivalent of Mg `=`equivalent of hydrogen.
`therefore ("weight of Al")/( "eq. wt. of Al") + ("weight of Mg")/("eq. wt. of Mg")="eq. of hydrogen" = ("volume of hydrogen (NTP)")/("equivalent volume of hydrogen")`.
Let the weight of `AI` be `x` g.
`therefore (x)/( 27//3) + (1-x)/( 24//2) = (1.104)/( 11.2) " "(E_("AI") = (27)/(3) )`
`(E_("Mg") = (24)/(2) )`
` x = 0.55`
`(1-x) = 0.45`.
Thus, weight of `"AI" = 0.55 g`
and weight of `"Mg" = 0.45 g`.
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