What mass of \(\mathrm{Na}_{2} \mathrm{CrO}_{4}\) is required to precipitate all of the silver ions from 75.0 mL of a 0.100-M solution of \(\mathrm{AgNO}_{3} ?\)

We have the precipitation reaction between sodium chromate (\(\mathrm{Na}_{2} \mathrm{CrO}_{4}\)) and silver nitrate (\(\mathrm{AgNO}_{3}\)). In this reaction, the silver ions will form a precipitate with chromate ions, resulting in the formation of silver chromate (\(\mathrm{Ag}_{2} \mathrm{CrO}_{4}\)). The balanced chemical equation for this reaction is: $$2\,\mathrm{AgNO}_{3} \,+\, \mathrm{Na}_{2}\mathrm{CrO}_{4} \,\longrightarrow\, \mathrm{Ag}_{2}\mathrm{CrO}_{4} \,+\, 2\,\mathrm{NaNO}_{3}$$

Next, we will determine the moles of silver ions in the \(\mathrm{AgNO}_{3}\) solution. To do this, we use the volume and concentration of the solution. Moles = Volume × Concentration The volume of the solution is 75.0 mL, which we must convert to liters. 1 L = 1000 mL, so we have: $$\text{Volume} = \frac{75.0\,\text{mL}}{1000\,\text{mL/L}} = 0.0750 \,\text{L}$$ Now, we can calculate the moles of \(\mathrm{AgNO}_{3}\): Moles of \(\mathrm{AgNO}_{3}\) = 0.0750 L × 0.100 mol/L = 0.00750 mol

Using the stoichiometry from the balanced chemical equation, we can determine the moles of \(\mathrm{Na}_{2} \mathrm{CrO}_{4}\) needed to react with the moles of \(\mathrm{AgNO}_{3}\). From the balanced equation, we can see that one mole of \(\mathrm{Na}_{2} \mathrm{CrO}_{4}\) is required to react with two moles of \(\mathrm{AgNO}_{3}\). Therefore, we can set up the following proportion to find the moles of \(\mathrm{Na}_{2} \mathrm{CrO}_{4}\) required: $$\frac{\text{moles of }\mathrm{Na}_{2}\mathrm{CrO}_{4}}{0.00750\,\text{mol } \mathrm{AgNO}_{3}} = \frac{1\,\text{mol }\mathrm{Na}_{2}\mathrm{CrO}_{4}}{2\,\text{mol}\,\mathrm{AgNO}_{3}}$$ We can solve for the moles of \(\mathrm{Na}_{2}\mathrm{CrO}_{4}\): $$\text{moles of }\mathrm{Na}_{2}\mathrm{CrO}_{4} = 0.00750\,\text{mol }\mathrm{AgNO}_{3} × \frac{1\,\text{mol}\,\mathrm{Na}_{2}\mathrm{CrO}_{4}}{2\,\text{mol }\mathrm{AgNO}_{3}} = 0.00375\,\text{mol }\mathrm{Na}_{2}\mathrm{CrO}_{4}$$

Finally, we can determine the mass of \(\mathrm{Na}_{2} \mathrm{CrO}_{4}\) required to react with the moles of silver ions present. We will use the molar mass of \(\mathrm{Na}_{2} \mathrm{CrO}_{4}\) to convert moles to grams. $$\text{Mass of }\mathrm{Na}_{2}\mathrm{CrO}_{4} =\text{Moles }× \text{Molar mass}$$ The molar mass of \(\mathrm{Na}_{2} \mathrm{CrO}_{4}\) is: $$2\,\text{mol Na }(22.99\,\text{g/mol}) + 1\,\text{mol Cr }(51.99\,\text{g/mol}) + 4\,\text{mol O }(16.00\,\text{g/mol}) = 161.97\,\text{g/mol}$$ Now we can find the mass of \(\mathrm{Na}_{2} \mathrm{CrO}_{4}\) required: $$\text{Mass of }\mathrm{Na}_{2}\mathrm{CrO}_{4} = 0.00375\,\text{mol} × 161.97\,\text{g/mol} = 0.607\,\text{g}$$ So, a mass of 0.607 g of \(\mathrm{Na}_{2} \mathrm{CrO}_{4}\) is required to precipitate all of the silver ions from 75.0 mL of 0.100-M solution of \(\mathrm{AgNO}_{3}\).