Answer: 7.88 g of AgCl are produced from 9.3 g of [tex]AgNO_3[/tex]
Explanation:
To calculate the moles :
[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]
[tex]\text{Moles of} AgNO_3=\frac{9.3 g}{169.8g/mol}=0.055moles[/tex]
[tex]2AgNO_3+BaCl_2\rightarrow 2AgCl+Ba(NO_3)_2[/tex]
According to stoichiometry :
2 moles of [tex]AgNO_3[/tex] produce = 2 moles of [tex]AgCl[/tex]
Thus 0.055 moles of [tex]AgNO_3[/tex] will produce=[tex]\frac{2}{2}\times 0.055=0.055moles[/tex] of [tex]AgCl[/tex]
Mass of [tex]AgCl=moles\times {\text {Molar mass}}=0.055moles\times 143.3g/mol=7.88g[/tex]
Thus 7.88 g of AgCl are produced from 9.3 g of [tex]AgNO_3[/tex]
