Question

What mass of HCL, in grams, is required to react with 0.610 g of al(oh)3 ?

Answer 1

Answer: 0.8541 grams of HCl will be required.

Step-by-step explanation: Moles can be calculated by using the formula:

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

Given mass of
`Al(OH)_3` = 0.610 g

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

`\text{Number of moles}=(0.610g)/(78g/mol)`

Number of moles of
`Al(OH)_3` = 0.0078 moles

The reaction between
`Al(OH)_3` and HCl is a type of neutralization reaction because here acid and base are reacting to form an salt and also releases water.

Chemical equation for the above reaction follows:

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

By Stoichiometry,

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

So, 0.0078 moles of
`Al(OH)_3` will react with
`(3)/(1)* 0.0078` = 0.0234 moles

Mass of HCl is calculated by using the mole formula, we get

Molar mass of HCl = 36.5 g/mol

Putting values in the equation, we get

`0.0234moles=\frac{\text{Given mass}}{36.5g/mol}`

Mass of HCl required will be = 0.8541 grams