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

Step-by-step explanation:

According to avogadro's law, 1 mole of every substance weighs equal to molecular mass and contains avogadro's number
`6.023* 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}=(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 =
`(3)/(1)* 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`