Question

Answer the following for the reaction: 3AgNO3(aq)+Na3PO4(aq)→Ag3PO4(s)+3NaNO3(aq)

1. How many milliliters of a 0.255 MAgNO3 solution are required to react with 31.1 mL of 0.135 MNa3PO4 solution?
2. How many grams of silver phosphate are produced from the reaction of 23.0 mL of a 0.195 MAgNO3 solution and excess Na3PO4?

Answer 1

Answer:1) Volume of
`AgNO_3` required is 55.98 mL.

2) 0.62577 grams of
`Ag_3PO_4` is produced.

Step-by-step explanation:

`3AgNO_3(aq)+Na_3PO_4(aq)\rightarrow Ag_3PO_4(s)+3NaNO_3(aq)`

1) Molarity of
`AgNO_3,M_1=0.225 M`

Volume of
`AgNO_3.V_1=?`

Molarity of
`Na_3PO_4,M_2=0.135 M`

Volume of
`Na_3PO_4,V_2=31.1 mL=0.0311 L`

`Molarity=\frac{\text{number of moles}}{\text{volume of solution in liters}}`

`\text{number of moles }Na_3PO_4=M_2* V_2=0.135 mol/L* 0.0311 L=0.0041985 moles`

According to reaction, 1 mole of
`Na_3PO_4` reacts with 3 mole of
`AgNO_3`, then, 0.0041985 moles of
`Na_3PO_4` will react with:

`(3)/(1)* 0.0041985` moles of
`AgNO_3` that is 0.0125955 moles.

`M_1=0.225 M=\frac{\text{number of moles of }AgNO_3}{V_1}`

`V_1=(0.0125955 moles)/(0.225 M)=0.05598 L=55.98 mL`

Volume of
`AgNO_3` required is 55.98 mL.

2)

`Molarity=0.195 M=\frac{\text{number of moles}}{\text{volume of solution in liters}}`

Number of moles of
`AgNO_3=0.195* 0.023 L=0.004485 moles`

According to reaction, 3 moles of
`AgNO_3` gives 1 mole of
`Ag_3PO_4`, then 0.004485 moles of
`AgNO_3` will give:
`(1)/(3)* 0.004485` moles of
`Ag_3PO_4` that is 0.001495 moles.

Mass of
`Ag_3PO_4` =

Moles of
`Ag_3PO_4` × Molar Mass of
`Ag_3PO_4`

= 0.001495 moles × 418.58 g/mol = 0.62577 g

0.62577 grams of
`Ag_3PO_4` is produced.