Question

What mass of $\mathit{N}{\mathit{a}}_{\mathbf{2}}\mathit{C}\mathit{r}{\mathit{O}}_{\mathbf{4}}$is required to precipitate all of the silver ions from 75.0 mL of a 0.100 M solution of $\mathit{A}\mathit{g}\mathit{N}{\mathit{O}}_{\mathbf{3}}$?

Short answer

Themass of$N{a}_{2}Cr{O}_4$ is required to precipitate all of the silver ions from 75.0 mL of a 0.100 M solution of$AgN{O}_3$ is$0.6074g$.

Step-by-step solution

Definition

Precipitation is a chemical reaction that results in theformation of an insoluble solid product from two or more soluble chemicals. Theprecipitate is the solid substance that results from this action.

Determination of mass of Na2CrO4required for AgNO3 

Here the given and itschemical reaction between $N{a}_{2}Cr{O}_4$and $AgN{O}_3$as follows,

$N{a}_{2}Cr{O}_4+2AgN{O}_3\to 2NaN{O}_3\left(aq\right)+A{g}_{2}Cr{O}_4\left(s\right)\downarrow$

So one mole of$N{a}_{2}Cr{O}_4$ is required to precipitate two moles of silver nitrate. Number of moles of

$AgN{O}_3=\left(Molarity\right)\left(volume\right)$

$=\left(0.100M\right)\left(0.0751\right)$

$=0.0075moles$

Here thenumber of molesof$N{a}_{2}Cr{O}_4$ required$=\frac{0.0075}{2}$

$=0.00375moles$

Then Molar massof$N{a}_{2}Cr{O}_4=2\left(23\right)+51.996+4\left(6\right)$

$=161.996\raisebox{1ex}{$g$}\!\left/ \!\raisebox{-1ex}{$mol$}\right.$

Thuscompute the mass of $0.00375moles$of,$N{a}_{2}Cr{O}_4$

$=\left(0.00375\right)\left(161.996\right)$

$=0.6074g$

Therefore the massof $0.00375moles$of $N{a}_{2}Cr{O}_4$is.$0.6074g$