Question

0.9g al reacts with dil. Hcl to give h2. The volume of h2 evolved at stp is

Answer 1

A reaction between an acid and a metal always in the formation of a salt and the liberation of hydrogen gas.
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In this specific question, when aluminum reacts with HCl, this would be the resulting reaction:

2 Al(s) + 6 HCl(aq) = 2 AlCl₃(aq) + 3 H₂(g)

This is an example of a redox reaction.

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To calculate the volume of hydrogen liberated, first find the moles of the Al

mass = 0.9 g
molar mass = 27

moles  = mass/molar mass

moles of 0.9g Al = 0.9/27
                           =  0.033

The mole ratio in the equation is  Al:HCl: AlCl₃:H₂   = 2:6:2:3

That means 2 moles of Al liberates 3 moles of H₂ with 6 moles of HCl

the ratio is 2:3

If 2 moles of Al give 3 moles of H₂
Then 0.033 moles of Al will give   = 0.33 × 3/2
                                                      = 0.05 moles

Therefore 0.9g of Al ( which is 0.33 moles) will give H₂ with 0.05 moles

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Find volume of H₂ gas liberated (use ideal gas law)

Ideal gas law states that 1 mole of an ideal gas occupies exactly 22.4 Liters by volume.

If 1 mole  = 22.4 liters
Then 0.05 moles = 22.4 × 0.05/1
                            = 1.12 liters

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Therefore when 0.9 g of Al reacts with HCl acid the volume of hydrogen gas liberated is 1.12 liters
                 

Answer 2

Answer:2Al+6HCL-->2AlCl3+3H2

2Al=54g

3h2=67.2liter

.: 54g al gives 67.2 liter h2

0.9g al gives 67.2/54 ×0.9

=1.12liter

Explanation: