Answer: 172.2 g of [tex]AgCl[/tex] is formed by adding sufficient silver nitrate to react with 1500.0mL of 0.400M barium chloride solution.
Explanation:
To calculate the moles, we use the equation:
[tex]\text{Number of moles}={\text {Molarty}}\times {\text{Volume in L}}=0.400\times 1.5L=0.6moles[/tex]
[tex]BaCl_2+2AgNO_3\rightarrow 2AgCl+Ba(NO_3)_2[/tex]
According to stoichiometry:
1 mole of [tex]BaCl_2[/tex] produce = 2 moles of [tex]AgCl[/tex]
Thus 0.6 moles [tex]BaCl_2[/tex] will produce =[tex]\frac{2}{1}\times 0.6=1.2[/tex] moles of [tex]AgCl[/tex]
Mass of [tex]AgCl=moles\times {\text {molar mass}}=1.2mol\times 143.5g/mol=172.2g[/tex]
Thus 172.2 g of [tex]AgCl[/tex] is formed by adding sufficient silver nitrate to react with 1500.0mL of 0.400M barium chloride solution.
