An alloy of Aluminum and copper was treated with aqueous HCL . the Alu

1.	An alloy of Aluminum and copper was treated with aqueous HCL . the Aluminum dissolved according to reactionAl + 3H⁺ → Al³⁺ +H₂But the copper remained as pure metal a 3 mole sample of alloy gave 67.2 liter of H₂ Measured at STP . what is the mole of Cu in the alloy ?
Answer» $\underline{\underline{\maltese\:\:\textbf{\textsf{Question}}}}$ An ALLOY of Aluminum and copper was treated with aqueous HCL . The Aluminum dissolved according to reaction $\sf{Al + 3H^+ \rightarrow Al^{3+} +H_2}$ . But the copper remained as pure metal , a 3 mole sample of alloy gave 67.2 liter of H₂ Measured at STP. What is the mole of Cu in the alloy ? $\underline{\underline{\maltese\:\:\textbf{\textsf{Given}}}}$ An alloy of Aluminum and copper was treated with aqueous HCL The Aluminum dissolved according to reaction $\sf{Al + 3H^+ \rightarrow Al^{3+} +H_2}$ Copper remained as pure metal A 3 mole sample of alloy gave 67.2 liter of H₂ Measured at STP Mole of Cu in the alloy $\rightarrow\textsf{To Find}$ $\underline{\underline{\maltese\:\:\textbf{\textsf{Solution}}}}$ The GENERAL equation is as follows : $\sf{A <klux>L</klux>+3 H C l \rightarrow A l C l_{3}+3/2\: H_{2}}$ It can be expressed as : $\sf{A l+3 H^{+} \rightarrow A l^{+3}+3/2\: H_{2}}$ We know : At STP, one mole of any gas occupies a volume of 22.4 L It means moles of 67.2 L of gas = $\sf{67.2 \:L/ 22.4 \:L}$ = 3 mole of H₂ From given reaction, it is clear that : $\sf{\textsf{ 1.5 } \textsf{mole of } H_{2} }=1 \textsf{ mole of } A l}$ $\longrightarrow\sf{3 \textsf { mole of } H_{2} =\dfrac{1 \textsf { mole of } A l}{1.5 \textsf { mole of } H_{2}} \times 3 \textsf { mole of } H_{2}}$ Hence, Mole of Al in alloy = 2 mole Mole of Cu in alloy = 3 mole - 2 mole = 1 mole It means there is 1 mole of Cu in alloy

An alloy of Aluminum and copper was treated with aqueous HCL . the Aluminum dissolved according to reactionAl + 3H⁺ → Al³⁺ +H₂But the copper remained as pure metal a 3 mole sample of alloy gave 67.2 liter of H₂ Measured at STP . what is the mole of Cu in the alloy ?

Answer»

$\underline{\underline{\maltese\:\:\textbf{\textsf{Question}}}}$

An ALLOY of Aluminum and copper was treated with aqueous HCL . The Aluminum dissolved according to reaction $\sf{Al + 3H^+ \rightarrow Al^{3+} +H_2}$ . But the copper remained as pure metal , a 3 mole sample of alloy gave 67.2 liter of H₂ Measured at STP. What is the mole of Cu in the alloy ?

$\underline{\underline{\maltese\:\:\textbf{\textsf{Given}}}}$

An alloy of Aluminum and copper was treated with aqueous HCL
The Aluminum dissolved according to reaction $\sf{Al + 3H^+ \rightarrow Al^{3+} +H_2}$
Copper remained as pure metal
A 3 mole sample of alloy gave 67.2 liter of H₂ Measured at STP
Mole of Cu in the alloy $\rightarrow\textsf{To Find}$

$\underline{\underline{\maltese\:\:\textbf{\textsf{Solution}}}}$

The GENERAL equation is as follows : $\sf{A <klux>L</klux>+3 H C l \rightarrow A l C l_{3}+3/2\: H_{2}}$

It can be expressed as : $\sf{A l+3 H^{+} \rightarrow A l^{+3}+3/2\: H_{2}}$

We know : At STP, one mole of any gas occupies a volume of 22.4 L

It means moles of 67.2 L of gas = $\sf{67.2 \:L/ 22.4 \:L}$ = 3 mole of H₂

From given reaction, it is clear that : $\sf{\textsf{ 1.5 } \textsf{mole of } H_{2} }=1 \textsf{ mole of } A l}$

$\longrightarrow\sf{3 \textsf { mole of } H_{2} =\dfrac{1 \textsf { mole of } A l}{1.5 \textsf { mole of } H_{2}} \times 3 \textsf { mole of } H_{2}}$

Hence, Mole of Al in alloy = 2 mole

Mole of Cu in alloy = 3 mole - 2 mole = 1 mole

It means there is 1 mole of Cu in alloy

An alloy of Aluminum and copper was treated with aqueous HCL . the Aluminum dissolved according to reactionAl + 3H⁺ → Al³⁺ +H₂But the copper remained as pure metal a 3 mole sample of alloy gave 67.2 liter of H₂ Measured at STP . what is the mole of Cu in the alloy ?

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