Determine the formula mass of Pb(NO_3)_2 in g/mol and convert 0.250 g Pb(NO_3)_2

Question

Determine the formula mass of Pb(NO_3)_2 in g/mol and convert 0.250 g Pb(NO_3)_2 to moles Pb(NO_3)_2. Determine the formula mass of KI in g/mol and convert 0.250 g KI to moles KI. How many moles of Pb(NO_3)_2 are there in 10.0 mL of a 0.10 M solution? How many moles of KI are there in 10.0 mL of a 0.10 M solution? Now assume in an experiment we mix 10.0 mL of a 0.10 M Pb(NO_3)_2 with 10.0 mL of a 0.10 M KI. Compare the number of moles of each reactant from your calculations in Qs 4 and 5, and determine which reactant is limiting and which reactant is in excess. Show your work. Limiting Reactant: Excess Reactant:

Explanation / Answer

Ans 4) Here Volume = 10 ml

molarity = 0.10M

Now we know

Molarity = (Number of moles / Volume (in ml)) x 1000

Number of moles of Pb(NO3)2 = Molarity x volume(in ml) /1000

                                            =0.10 x10 /1000

                                            = 0.001

Ans 5) Here Volume = 10ml

Molarity =0.10M

Number of moles of KI = Molarity x volume(in ml) /1000

                                 = 0.10x10/1000

   =0.001

Ans 6) The balanced chemical reaction is

Pb(NO3)2 + 2KI -> PbI2 +2KNO3

Here There is 1mol of lead nitrate and 2mol of KI

Therefore Number of moles of KI

= 0.10 x0.1 x2mol KI

= 0.001 x 2

=0.002 mol of KI

Again , for 1mol of Pb(NO3)2

number of moles = 0.01 x0.1

   =0.001 mol Pb(NO3)2

Here we can see that lead nitrate produces less number of moles so it the limiting reagent

KI is the excess reagent with more number of moles.

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