Question

The following is a precipitation reaction: 2 K3PO4 (aq) + 3 Co(NO3)2 (aq) → Co3(PO4)2 (s) + 6 KNO3 (aq) What volume of 0.222 M K3PO4 (aq) in milliliters is needed to react with 36.48 mL of 0.250 M Co(NO3)2 (aq)?

Answer 1

Answer – We are given, [K₃PO₄] = 0.222 M , [Co(NO₃)₂] = 0.250 M , volume = 36.48 mL

2 K₃PO₄ (aq) + 3 Co(NO₃)₂ (aq) -----> Co₃(PO₄)₂ (s) + 6 KNO₃ (aq)

First we need calculate moles of Co(NO₃)₂

We know,

Moles of Co(NO₃)₂ = molarity of Co(NO₃)₂* volume (L)

                                = 0.250 M * 0.03648 L

                                = 0.00912 moles

From the balanced reaction –

3 moles of Co(NO₃)₂ = 2 moles of K₃PO₄

0.00912 moles of Co(NO₃)₂ = ?

= 0.00608 moles K₃PO₄

Now we are given the molarity of the K₃PO₄ and we calculated the moles of K₃PO₄, so using the molarity and volume we can calculate the volume

We know formula,

Molarity = moles / volume (L)

So, volume (L) = moles / molarity

                          = 0.00608 moles / 0.222 M

                           = 0.0274 L

Now we need to convert the volume L to mL and we know

1 L = 1000 mL

So, 0.0274 L = ?

= 27.4 mL

So, 27.4 mL of 0.222 M K₃PO₄ (aq) is needed to react with 36.48 mL of 0.250 M Co(NO₃)₂ (aq).