Answer :
Answer : The mass of [tex]Ag_2CrO_4[/tex] precipitate will be, 12.4 grams.
Explanation :
First we have to calculate the moles of [tex]AgNO_3[/tex]
[tex]\text{Moles of }AgNO_3=\text{Concentration of }AgNO_3\times \text{Volume of solution}[/tex]
[tex]\text{Moles of }AgNO_3=0.500M\times 0.150L=0.075mol[/tex]
Now we have to calculate the moles of [tex]Ag_2CrO_4[/tex]
The balanced chemical reaction is:
[tex]2AgNO_3(aq)+K_2CrO_4(aq)\rightarrow Ag_2CrO_4(s)+2KNO_3(aq)[/tex]
From then balanced chemical reaction we conclude that,
As, 2 moles of [tex]AgNO_3[/tex] react to give 1 mole of [tex]Ag_2CrO_4[/tex]
So, 0.075 moles of [tex]AgNO_3[/tex] react to give [tex]\frac{0.075}{2}=0.0375[/tex] mole of [tex]Ag_2CrO_4[/tex]
Now we have to calculate the mass of [tex]Ag_2CrO_4[/tex]
[tex]\text{ Mass of }Ag_2CrO_4=\text{ Moles of }Ag_2CrO_4\times \text{ Molar mass of }Ag_2CrO_4[/tex]
Molar mass of [tex]Ag_2CrO_4[/tex] = 331.73 g/mole
[tex]\text{ Mass of }Ag_2CrO_4=(0.0375moles)\times (331.73g/mole)=12.4g[/tex]
Therefore, the mass of [tex]Ag_2CrO_4[/tex] precipitate will be, 12.4 grams.