Question

Assume that 13.5 g solid aluminum (Al) react with HCl to produce solid aluminum chloride (AlCl3) salt and gaseous hydrogen (H2) at standard temperature and pressure. How many moles of Al react? mol How many moles of H2 are produced? mol How many liters of H2 are produced? L

Answer 1

hello

The balance chemical reaction is 2Al + 6HCl -à 2AlCl3 + 3H2 Assuming excess amount of HCl because it is not given Mole Al reacted = 13.5 g ( 1 mole/ 27 g) = 0.5 mole Al
Mole H2 = 0.5 mol Al ( 3 mole H2 / 2 mole Al) = 0.75 mole H2
since at STP then 1 mole gas occupies 22.414 L volume H2 = 0.75 mole (22.414 L / mol) = 16.8105 L H2
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Answer 2

Answer : The moles of Al is, 0.50 mole.

The moles of
`H_2` is, 0.75 mole.

The volume of
`H_2` is, 16.8 liters.

Explanation : Given,

Mass of aluminum = 13.5 g

Molar mass of aluminum = 26.98 g/mole

First we have to calculate the moles of aluminum.

`\text{Moles of }Al=\frac{\text{Mass of }Al}{\text{Molar mass of }Al}=(13.5g)/(26.98g/mole)=0.50moles`

The moles of Al is, 0.50 mole.

Now we have to calculate the moles of
`H_2`.

The balanced chemical reaction will be,

`2Al(s)+6HCl(l)\rightarrow 2AlCl_3(s)+3H_2(g)`

From the balanced reaction, we conclude that

As, 2 moles of Al react to give 3 moles of
`H_2`

So, 0.50 moles of Al react to give
`(3)/(2)* 0.50=0.75` moles of
`H_2`

The moles of
`H_2` is, 0.75 mole.

Now we have to calculate the volume of
`H_2`.

At STP,

As, 1 mole of
`H_2` contains 22.4 L volume of
`H_2`

So, 0.75 mole of
`H_2` contains
`0.75* 22.4=16.8L` volume of
`H_2`

The volume of
`H_2` is, 16.8 liters.