Question

what volume of 0.305 M AgNO3 is required to react exactly with 155.0 ml of 0.274 Monday Na2SO4 solution?

Answer 1

Number of moles of Na₂SO₄ , n = Molarity x volume in L

                                               = 0.274 M x 155.0 mL x 10^-3 L/mL

                                                = 0.0425 moles

The balanced equation is :   Na₂SO₄(aq) + 2AgNO₃(aq) ----> Ag₂SO₄(s) + 2 NaNO₃(aq)

From the balanced equation,

1 mole of Na₂SO₄ reacts with 2 moles of AgNO₃

0.0425 moles of Na₂SO₄ reacts with 2x0.0425 = 0.0850 moles of AgNO₃

We know that Molarity , M = number of moles / volume of the solution in L

So the volume of AgNO3 reacted is , V = number of moles / molarity

                                                         = 0.0850 mol / 0.305 M

                                                        = 0.2787 L

                                                        = 278.7 mL

Therefore the volume of AgNO3 required is 278.7 mL