Question

How many L of 1.80 M HCl are required to react completely with 12.0 g of Al(OH)3 ? Al(OH)3 + 3HCl AlCl3 + 3H2O

Answer 1

Answer: 0.257 L

Explanation:

According to avogadro's law, 1 mole of every substance weighs equal to molecular mass and contains avogadro's number $6.023\times 10^{23}$ of particles.

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

given mass of $Al(OH)_3$ = 12.0 g

Molar mass of $Al(OH)_3$ =  78 g/mol

Putting in the values we get:

$\text{Number of moles}=\frac{12.0g}{78g/mol}=0.154moles$

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

According to stoichiometry:

1 mole of $Al(OH)_3$ reacts with 3 moles of $HCl$

Thus 0.154 moles of $Al(OH)_3$ reacts with =$\frac{3}{1}\times 0.154=0.462moles$ of $HCl$

To calculate the volume for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$  

$1.80M=\frac{0.462}{\text{Volume of solution in L}}$

${\text{Volume of solution in L}}=0.257$

Thus 0.257 L of 1.80 M HCl are required to react completely with 12.0 g of $Al(OH)_3$