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* \text{Volume of solution}`

`\text{Moles of }AgNO_3=0.500M* 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
`(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* \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)* (331.73g/mole)=12.4g`

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