What volume of 0.105 M Na2S solution will exactly react with 25.0 mL of 0.140 M aGNO3 solution?

Question

2 AGNO (aq) + NA S(aq) ---> Ag S(s) +2NaNO
............3 .............2 ................2 ................3

Answer 1

Find the moles of AgNO3:

n = C*V, n = 0.14*25x10^(-3) mol

According to the chem. equation:

2AgNO3 + Na2S --> 2NaNO3 + Ag2S

2 moles of AgNO3 reacts fully with 1 mole of Na2S, so
0.14*25x10^(-3) moles of AgNO3 reacts with 0.07*25x10^(-3) moles of Na2S

Now you can find the volume of the Na2S solution:

n = C*V => V = n/C = 0.07*25x10^(-3)/0.105 = 0.05/3 L or

V = 16.7 mL (approx)

Answer 2

From the equation:

2AgNO3 + Na2S ----> Ag2S + 2NaNO

First, you need to get moles of AgNO3:

(25.0 mL)(1L/1000mL) = 0.025 L AgNO3

(molarity) * (volume) = moles
(0.140 mol AgNO3/1L AgNO3)*(0.025 L AgNO3) = 0.0035 mol

AgNO3

the mole ratio from the equation says 2 moles of AgNO3 reacts with 1 mol Na2S, so...

(0.0035 mol AgNO3)(1 mol Na2S/2mol AgNO3) = 0.00175 mol

Na2S.

You're given the molarity of Na2S, so you can calculate the volume that will react exactly with the 25.0 mL of AgNO3.

(1 L Na2S/0.105 mol Na2S)*(0.00175 mol Na2S)(1000 mL/1L)=

16.7 mL Na2S :)