How many milliliters of 0.260 m na2s are needed to react with 40.00 ml of 0.315 m agno3? na2s(aq) + 2 agno3(aq) → 2 nano3(aq) + ag2s(s)?

the balanced equation for the reaction is as follows ;
                                                                   
Na₂S + 2AgNO₃ ---> 2NaNO₃ + Ag₂S
                                                                   
stoichiometry of Na₂S to AgNO₃ is 1:2
                                                                     
number of AgNO₃ moles reacted - 0.315 mol/L x 0.04000L = 0.0126 mol according to molar ratio of 1:2
                                                                   
number of Na₂S moles required are - 1/2 x number of AgNO3 moles reacted Na₂S moles = 0.0126 mol /2 = 0.00630 mol
                                                                     
molarity of Na₂S - 0.260 M
                                                                   
there are 0.260 mol in 1 L
                                                                     
therefore 0.00630 mol are in - 0.00630 mol / 0.260 mol/L
                                                                   
volume of Na₂S required = 0.0242 L
                                                                   
volume of Na₂S required = 24.4 mL

The answer is 2.42 ml (milliliter)

Determine the balanced equation:

Balanced equation = NA2S (aq) + 2 AGNO3 (aq) > 2NANO3(aq) + AG2(s)

Let:

Number of moles of AGNO3 = AGNO3 * vol. of AGNO3

Substitute values:

number of moles = 0.315 * 0.04

number of moles = 0.0126 AGNO3

(For NA2s)

Let:

Number of moles of NA2s = 1/2 (number of moles of AGNO3)

number of moles of NA2s = 0.0063

volume of NA2s needed = number of mole of NA2s / NA2s

For the volume:

Volume = 0.0063 / 0.260

thus,

V = 0.02423 L (should be converted to milliliter)

Therefore the Volume in ml = 2.42 ml (milliliter)