Synthesis of Lead(II) Chromate – Flashcards
Unlock all answers in this set
Unlock answersquestion
when 2 chemicals react , sometime result in the formation of this
answer
Precipitate
question
the reaction for this expreiment (calculation)
answer
K2CrO4(aq) +Pb(NO3)2 ->PbCrO4(s) + 2KNO3(aq)
question
in all experiments , the number of Moles of a product formed will depend on the number of moles in the reactant
answer
limiting reactant
question
In the quantity of the reactant is such that the ratio is not 1to1 , the excess of reactant will not react.
answer
ratio
question
limiting reactant in this experiment
answer
lead nitrate
question
= (molarity of solution, mol L -1) * (volume of solution, L)
answer
number of moles of reactant
question
(0.0900 mol L(-1))(0.0500L) = 4.50*10(-3)
answer
# of Moles of K2CrO4
question
(0.100 mol L(-1) )(0.0500 L) = 5.00 *10(-3) or 1.4286g/323.22g mol(-1) = 4.42*10(-3)
answer
number of moles of Pb(N03)2
question
when one is in excess , the other is the limiting reagent. for this experiment lead chormate is in excess and potassium is the limiting reagent
answer
limiting reagent
question
reactant - limits the amount of combining, (reaction) reagent - limits the amount that is left after reaction
answer
difference between reactant and reagent
question
(#of moles od PbCr04 formed/theorectical # of PbCrO4) *100 or (4.42 *10(-3)/4.50 *10(-3)) *100=98.2%
answer
Percent yield of lead(II) chromate calculation
question
in the lab we used a heat lamp versus an oven ,
answer
what difference from lab and book
