Question

9. In the reaction of sodium hydroxide with chlorine gas, sodium chloride, sodium hypochlorite, and water a

Answer 1

Answer: 51.4 g of NaClO will be produced.

Explanation:

To calculate the moles :

$\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}$    

$\text{Moles of} Cl_2=\frac{48.9g}{71g/mol}=0.69moles$

$\text{Moles of} NaOH=\frac{54.2g}{40g/mol}=1.4moles$

$2NaOH+Cl_2\rightarrow NaCl+NaClO+H_2O$

According to stoichiometry :

1 moles of $Cl_2$ require = 2 moles of $NaOH$

Thus 0.69 moles of $Cl_2$ will require=$\frac{2}{1}\times 0.69=1.38moles$  of $NaOH$

Thus $Cl_2$ is the limiting reagent as it limits the formation of product and $NaOH$ is the excess reagent.

As 1 mole of $Cl_2$ give = 1 mole of $NaClO$

Thus 0.69 moles of $Cl_2$ give =$\frac{1}{1}\times 0.69=0.69moles$  of $NaClO$

Mass of $NaClO=moles\times {\text {Molar mass}}=0.69moles\times 74.5g/mol=51.4g$

Thus 51.4 g of NaClO will be produced