Question

Calculate the number of grams of aluminum chloride and water formed when 0.500 g of aluminum hydroxide react with stomach acid.

Answer 1

Answer : The mass of
`AlCl_3` and
`H_2O` produced will be, 0.852 and 0.346 grams respectively.

Explanation : Given,

Mass of
`Al(OH)_3` = 0.500 g

Molar mass of
`Al(OH)_3` = 78 g/mole

Molar mass of
`AlCl_3` = 133 g/mole

Molar mass of
`H_2O` = 18 g/mole

First we have to calculate the moles of
`Al(OH)_3`.

`\text{Moles of }Al(OH)_3=\frac{\text{Mass of }Al(OH)_3}{\text{Molar mass of }Al(OH)_3}=(0.500g)/(78g/mole)=0.00641moles`

Now we have to calculate the moles of
`AlCl_3` and
`H_2O`.

The balanced chemical reaction is,

`Al(OH)_3+3HCl\rightarrow AlCl_3+3H_2O`

From the balanced reaction we conclude that

As, 1 mole of
`Al(OH)_3` react to give 1 mole of
`AlCl_3`

So, 0.00641 mole of
`Al(OH)_3` react to give 0.00641 mole of
`AlCl_3`

And,

As, 1 mole of
`Al(OH)_3` react to give 3 mole of
`H_2O`

So, 0.00641 mole of
`Al(OH)_3` react to give
`3* 0.00641=0.01923` mole of
`H_2O`

Now we have to calculate the mass of
`AlCl_3` and
`H_2O`.

`\text{Mass of }AlCl_3=\text{Moles of }AlCl_3* \text{Molar mass of }H_2O`

`\text{Mass of }AlCl_3=(0.00641mole)* (133g/mole)=0.852g`

and,

`\text{Mass of }H_2O=\text{Moles of }H_2O* \text{Molar mass of }H_2O`

`\text{Mass of }H_2O=(0.01923mole)* (18g/mole)=0.346g`

Therefore, the mass of
`AlCl_3` and
`H_2O` produced will be, 0.852 and 0.346 grams respectively.