1.00 × 10-3 moles of Ag + and 1.00 × 10-3 moles of CrO 42- react together to form solid Ag 2 CrO 4 Calculate the amount of Ag 2 CrO 4 formed. ( Ag 2 CrO 4=331.73 g mol -1)

Question

1.00 × 10-3 moles of Ag + and 1.00 × 10-3 moles of CrO 42- react together to form solid Ag 2 CrO 4 Calculate the amount of Ag 2 CrO 4 formed. ( Ag 2 CrO 4=331.73 g mol -1) (A) 0.268 g (B) 0.166 g (C) 0.212 g (D) 1.66 g

Tardigrade Admin · Accepted Answer

The reaction is 2 Ag ++ CrO 42- → Ag 2 CrO 4 Using the limiting reagent concept, amount of Ag 2 CrO 4 =0.5 × 10-3 × 331.73=0.166 g

Q. $1.00 \times 1 0^{- 3}$ moles of $A g^{+}$ and $1.00 \times 1 0^{- 3}$ moles of $C r O_{4}^{2 -}$ react together to form solid $A g_{2} C r O_{4}$ Calculate the amount of $A g_{2} C r O_{4}$ formed. $(A g_{2} C r O_{4} = 331.73 g m o l^{- 1})$

$0.268 g$

$0.166 g$

$0.212 g$

$1.66 g$

The reaction is $2 A g^{+} + C r O_{4}^{2 -} \to A g_{2} C r O_{4}$

Using the limiting reagent concept, amount of $A g_{2} C r O_{4}$

$= 0.5 \times 1 0^{- 3} \times 331.73 = 0.166 g$

Q. 1.00×10−3 moles of Ag+ and 1.00×10−3 moles of CrO42−​ react together to form solid Ag2​CrO4​ Calculate the amount of Ag2​CrO4​ formed. (Ag2​CrO4​=331.73gmol−1)

0.268g

0.166g

0.212g

1.66g