Question

39g of an alloy of Al and Mg when heated with excess of dilute HCl forms MgCl2 and AlCl3 and evolves Hydrogen gas which has 44.8L at STP. Find the mass of Mg and Al .​

Answer 1

$\red{\bigstar}\underline{\underline{\textsf{\textbf{ Given :- }}}}$

• 39g of an alloy of Al and Mg when heated with excess of dilute HCl forms MgCl₂ and AlCl₃ .
• The volume of Hydrogen gas evolved at STP is 44.8 L .

$\red{\bigstar}\underline{\underline{\textsf{\textbf{ To Find :- }}}}$

• The mass of Mg and Al .

$\red{\bigstar}\underline{\underline{\textsf{\textbf{ Solution :- }}}}$

Let us take the mass of Magnesium be x grams and the mass of Aluminium be ( 39 - x ) grams .

As , we know that ,

$\sf\longrightarrow Molecular\ wt \ of \ Mg \ = 24g$

$\sf\longrightarrow Molecular\ wt \ of \ Al \ = 27g$

Now here , we can find the number of moles as ,

$\sf\longrightarrow n_{(Mg)}= \dfrac{ Given \ wt.}{Molecular\ wt.} =\red{\dfrac{ x}{24}}$

$\sf\longrightarrow n_{(Al)}= \dfrac{ Given \ wt.}{Molecular\ wt.} =\red{\dfrac{ (39-x)}{27}}$

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• Now here let's find out the number of moles of Hydrogen gas . Since the gas is at STP , the volume of one mole of gas will be 22.4 L . Since the Hydrogen is 44.8 L . Therefore the number of moles will be ,

$\sf\longrightarrow n_{(H_2)}= \dfrac{ Volume}{22.4 \ L }= \dfrac{44.8L}{22.4L}=\red{2}$

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First Reaction will be ,

$\sf\dashrightarrow \purple{1}Mg + \purple{2}HCl \to MgCl_2 + \purple{1}H_2$

• Thus we can see that , the mole ratio of Mg and Hydrogen is 1:1 . Since the number of moles of Mg was x/24 . Therefore the number of moles of Hydrogen is ,

$\sf\longrightarrow n_{(H_2)_{1st\ Reaction} }=\red{\dfrac{x}{24}}$

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Second Reaction will be ,

$\sf\dashrightarrow \purple{1}Al + \purple{3}HCl \to AlCl_3 + \purple{\dfrac{3}{2}}H_2$

• Therefore the number of moles of Hydrogen released from second Reaction will be ,

$\sf\longrightarrow n_{(H_2)_{2nd\ Reaction} }=\red{\dfrac{3}{2}\bigg(\dfrac{39-x}{27}\bigg)}$

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• And by Question , total number of moles found was 2 . Therefore ,

$\sf\longrightarrow \dfrac{x}{24}+\dfrac{3}{2}\bigg(\dfrac{39-x}{27} \bigg) = 2 \\\\\\\sf\longrightarrow \dfrac{x}{24} + \dfrac{117-3x}{54}=2 \\\\\\\sf\longrightarrow \dfrac{ 9x -12x + 468 }{216}= 2 \\\\\\\sf\longrightarrow -3x +468 = 2*216 \\\\\\\sf\longrightarrow -3x = 432-468\\\\\\\sf\longrightarrow -3x = -36 \\\\\\\sf\longrightarrow \boxed{\pink{\frak{ \quad x \quad =\quad 12 }}}$

Therefore ,

$\underset{\blue{\sf Required\ Answer }}{\underbrace{\boxed{\begin{array}{c}\sf\longrightarrow Given \ wt .\ of \ Mg \ = \pink{12g }\\\\\\\sf\longrightarrow Given\ wt \ of \ Al \ =\pink{ 27g} \end{array}}}}$

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