Respuesta :
Answer : The mass of [tex]AlCl_3[/tex] and [tex]H_2O[/tex] produced will be, 0.852 and 0.346 grams respectively.
Explanation : Given,
Mass of [tex]Al(OH)_3[/tex] = 0.500 g
Molar mass of [tex]Al(OH)_3[/tex] = 78 g/mole
Molar mass of [tex]AlCl_3[/tex] = 133 g/mole
Molar mass of [tex]H_2O[/tex] = 18 g/mole
First we have to calculate the moles of [tex]Al(OH)_3[/tex].
[tex]\text{Moles of }Al(OH)_3=\frac{\text{Mass of }Al(OH)_3}{\text{Molar mass of }Al(OH)_3}=\frac{0.500g}{78g/mole}=0.00641moles[/tex]
Now we have to calculate the moles of [tex]AlCl_3[/tex] and [tex]H_2O[/tex].
The balanced chemical reaction is,
[tex]Al(OH)_3+3HCl\rightarrow AlCl_3+3H_2O[/tex]
From the balanced reaction we conclude that
As, 1 mole of [tex]Al(OH)_3[/tex] react to give 1 mole of [tex]AlCl_3[/tex]
So, 0.00641 mole of [tex]Al(OH)_3[/tex] react to give 0.00641 mole of [tex]AlCl_3[/tex]
And,
As, 1 mole of [tex]Al(OH)_3[/tex] react to give 3 mole of [tex]H_2O[/tex]
So, 0.00641 mole of [tex]Al(OH)_3[/tex] react to give [tex]3\times 0.00641=0.01923[/tex] mole of [tex]H_2O[/tex]
Now we have to calculate the mass of [tex]AlCl_3[/tex] and [tex]H_2O[/tex].
[tex]\text{Mass of }AlCl_3=\text{Moles of }AlCl_3\times \text{Molar mass of }H_2O[/tex]
[tex]\text{Mass of }AlCl_3=(0.00641mole)\times (133g/mole)=0.852g[/tex]
and,
[tex]\text{Mass of }H_2O=\text{Moles of }H_2O\times \text{Molar mass of }H_2O[/tex]
[tex]\text{Mass of }H_2O=(0.01923mole)\times (18g/mole)=0.346g[/tex]
Therefore, the mass of [tex]AlCl_3[/tex] and [tex]H_2O[/tex] produced will be, 0.852 and 0.346 grams respectively.
