Chemistry, asked by Anonymous, 3 months ago

An alloy of Aluminum and copper was treated with aqueous HCL . the Aluminum dissolved according to reaction
Al + 3H⁺ → Al³⁺ +H₂
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 ?

Answers

Answered by RockingStarPratheek
34

\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 l+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


Anonymous: Great :)
Answered by ItzMeMukku
2

Explanation:

\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}Al+3H

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}Al+3H

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}

→To Find

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

The general equation is as follows : \sf{A l+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}67.2L/22.4L = 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.

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