# In the reaction,$\dpi{100} 2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}$ 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 .

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.

N

3

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