Question

$1.00 \times 10^{-3}$ moles of $Ag ^{+}$ and $1.00 \times 10^{-3}$ moles of $CrO _{4}^{2-}$ react together to form solid $Ag _{2} CrO _{4} $ Calculate the amount of $Ag _{2} CrO _{4}$ formed. $\left( Ag _{2} CrO _{4}=331.73 \,g \,mol ^{-1}\right)$

• A. $0.268\, g$
• B. $0.166 \,g$
• C. $0.212 \,g$
• D. $1.66 \,g$

Answer 1

Answer: (B)

The reaction is $2 Ag ^{+}+ CrO _{4}^{2-} \to Ag _{2} CrO _{4}$

Using the limiting reagent concept, amount of $Ag _{2} CrO _{4}$

$=0.5 \times 10^{-3} \times 331.73=0.166\, g$