Question

45) 2 AgNO 3 (aq) + K 2 CrO 4 (aq) -> Ag 2 CrO 4 (s) + 2 KNO 3 (aq) How many grams of silver chromate will precipitate when 150.0 mL of 0.500 M silver nitrate is added to excess potassium chromate?

Answer 1

Answer : The mass of $Ag_2CrO_4$ precipitate will be, 12.4 grams.

Explanation :

First we have to calculate the moles of $AgNO_3$

$\text{Moles of }AgNO_3=\text{Concentration of }AgNO_3\times \text{Volume of solution}$

$\text{Moles of }AgNO_3=0.500M\times 0.150L=0.075mol$

Now we have to calculate the moles of $Ag_2CrO_4$

The balanced chemical reaction is:

$2AgNO_3(aq)+K_2CrO_4(aq)\rightarrow Ag_2CrO_4(s)+2KNO_3(aq)$

From then balanced chemical reaction we conclude that,

As, 2 moles of $AgNO_3$ react to give 1 mole of $Ag_2CrO_4$

So, 0.075 moles of $AgNO_3$ react to give $\frac{0.075}{2}=0.0375$ mole of $Ag_2CrO_4$

Now we have to calculate the mass of $Ag_2CrO_4$

$\text{ Mass of }Ag_2CrO_4=\text{ Moles of }Ag_2CrO_4\times \text{ Molar mass of }Ag_2CrO_4$

Molar mass of $Ag_2CrO_4$ = 331.73 g/mole

$\text{ Mass of }Ag_2CrO_4=(0.0375moles)\times (331.73g/mole)=12.4g$

Therefore, the mass of $Ag_2CrO_4$ precipitate will be, 12.4 grams.