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In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

 Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced
Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

 Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

Answers (3)

best_answer

As we learnt in

Number of Moles of a gas at STP -

no of moles of a gas at STP= given volume of gas / 22.4 liter

-

 

 In given problem 2 mole Al produce 3 mole H2(g)

Hence 1 mole Al give \frac{3}{2} mole H2(g)

\therefore  Volume of 1 mole H2 at STP is 22.4 L

\therefore Volume of \frac{3}{2} mole H2 at STP =\frac{3}{2}\times 22.4=33.6 L\ H_{2}

Ans. option 3


Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

This is an incorrect option.

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

This is an incorrect option.

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

This is the correct option.

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

This is an incorrect option.

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