Answer: D. 239 g of [tex]Br_2[/tex]
Explanation:
To calculate the moles :
[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]
[tex]\text{Moles of} KBr=\frac{356g}{119g/mol}=2.99moles[/tex]
The balanced chemical reaction is:
[tex]Cl_2+2KBr\rightarrow Br_2+2KCl[/tex]
According to stoichiometry :
2 moles of [tex]KBr[/tex] produce = 1 mole of [tex]Br_2[/tex]
Thus 2.99 moles of [tex]KBr[/tex] produce =[tex]\frac{1}{2}\times 2.99=1.495moles[/tex] of [tex]Br_2[/tex]
Mass of [tex]Br_2=moles\times {\text {Molar mass}}=1.495moles\times 159.8g/mol=239g[/tex]
Thus 239 g of [tex]Br_2[/tex] will be produced from 356 g of KBr.
