Answer:
10.7 g AgNO₃
Explanation:
First, we can set up the reaction equation:
[tex]\rm 3\ AgNO_3_{(aq)} + Na_3PO_4_{(aq)} \to Ag_3PO_4_{(s)} + 3\ NaNO_3_{(aq)}[/tex]
We can identify that this is a double replacement reaction that creates a precipitate (an insoluble solid) because:
Next, we can find the number of moles of Na₃PO₄ that we have by multiplying its volume (converted to liters) by its molarity:
[tex]\rm (0.02002\ L)\!\left(\dfrac{1.05\ mole}{1\ L}\right) = 0.021021 \ mole\ Na_3PO_4[/tex]
Then, we can find how many moles of AgNO₃ this would react with by multiplying by the stoichiometric ratio from the reaction equation:
[tex]\rm (0.021021\text{ mole}\ Na_3PO_4)\!\left(\dfrac{3\ moles\ AgNO_3}{1\ mole\ Na_3PO_4}\right) = 0.063063\ mole\ AgNO_3[/tex]
Finally, we can convert moles to grams using the molar mass of AgNO₃:
[tex]\displaystyle M_{\rm AgNO_3} = \dfrac{(107.87 + 14.01+3(16.00))\text{ g}}{1\text{ mole}} = 169.88\text{ g/mole}[/tex]
[tex]m=n\cdot M[/tex]
[tex]m = (0.063063\text{ mole AgNO}_3)\!\left(\dfrac{169.88\text{ g}}{1\text{ mole}}\right) = \boxed{10.7\text{ g AgNO}_3}[/tex]
So, we need 10.7 g of silver nitrate to completely react with 20.02 mL of 1.05 M sodium phosphate.
