Respuesta :
Answer: 51.4 g of NaClO will be produced.
Explanation:
To calculate the moles :
[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]
[tex]\text{Moles of} Cl_2=\frac{48.9g}{71g/mol}=0.69moles[/tex]
[tex]\text{Moles of} NaOH=\frac{54.2g}{40g/mol}=1.4moles[/tex]
[tex]2NaOH+Cl_2\rightarrow NaCl+NaClO+H_2O[/tex]
According to stoichiometry :
1 moles of [tex]Cl_2[/tex] require = 2 moles of [tex]NaOH[/tex]
Thus 0.69 moles of [tex]Cl_2[/tex] will require=[tex]\frac{2}{1}\times 0.69=1.38moles[/tex] of [tex]NaOH[/tex]
Thus [tex]Cl_2[/tex] is the limiting reagent as it limits the formation of product and [tex]NaOH[/tex] is the excess reagent.
As 1 mole of [tex]Cl_2[/tex] give = 1 mole of [tex]NaClO[/tex]
Thus 0.69 moles of [tex]Cl_2[/tex] give =[tex]\frac{1}{1}\times 0.69=0.69moles[/tex] of [tex]NaClO[/tex]
Mass of [tex]NaClO=moles\times {\text {Molar mass}}=0.69moles\times 74.5g/mol=51.4g[/tex]
Thus 51.4 g of NaClO will be produced
