Question

How many gram slaked lime is required to prepare 143 g bleaching powder?​

Answer 1

Answer: The mass of slaked lime required is 74.1 grams

Explanation:

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

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$     .....(1)

Given mass of bleaching powder = 143 g

Molar mass of bleaching powder = 143 g/mol

Putting values in equation 1, we get:

$\text{Moles of bleaching powder}=\frac{143g}{143g/mol}=1mol$

The chemical reaction for the formation of bleaching powder from slaked lime follows:

$Ca(OH)_2+Cl_2\rightarrow CaOCl_2+H_2O$

By Stoichiometry of the reaction:

1 mole of bleaching powder is produced from 1 mole of slaked lime

Now, calculating the mass of slaked lime by using equation 1, we get:

Molar mass of slaked lime = 74.1 g/mol

Moles of slaked lime = 0.042 moles

Putting values in equation 1, we get:

$1mol=\frac{\text{Mass of slaked lime}}{74.1g/mol}\\\\\text{Mass of slaked lime}=(1mol\times 74.1g/mol)=74.1g$

Hence, the mass of slaked lime required is 74.1 grams