A student requires 50 mL of a KI solution to completely react it with the Pb(NO3)2 in 50 mL of 2.00 M Pb(NO3)2. The balanced equation for the reaction is as follows.

Pb(NO3)2 + 2KI mc026-1.jpg PbI2 + 2KNO3

If 20.0 M KI is the only stock solution available, which correctly explains how the student should prepare the 50 mL of KI solution?

Question

A student requires 50 mL of a KI solution to completely react it with the Pb(NO3)2 in 50 mL of 2.00 M Pb(NO3)2. The balanced equation for the reaction is as follows.

Pb(NO3)2 + 2KI mc026-1.jpg PbI2 + 2KNO3

If 20.0 M KI is the only stock solution available, which correctly explains how the student should prepare the 50 mL of KI solution?

anonymous · Answer

@robtobey @Mertsj Help please

mertsj · Answer

The 50 ml of 2 M lead nitrate contains 50/1000 (2) moles of lead nitrate or 1/10 mole of lead nitrate.
According to the balanced equation you need twice as many moles of potassium iodide or 1/5 mole or .2 mole of potassium iodide.  It is to be contained in 50 ml of solution.
The 20 M KI has 20 moles per 1000 ml
20/1000=.2/x
20x=200
x=10
So 10 ml of the 20 M KI contains .2 moles KI 
The student should take 10 ml of the 20 M solution and dilute it to 50 ml