A student reacted 0.500g of Pb(NO3)2 with 0.750g of KI according to the reaction,

Pb(NO3)2(aq) + 2KI(aq)=PbI2(s) + 2KNO3(aq)

A) How many moles of Pb(NO3)2 were used?
B) How many moles of KI were used?
C) How many moles of PbI2 would form,
based on the moles of Pb(NO3)2 used?
D) How many moles of PBI2 would form,
based on the moles of KI used?
E) Which is the limiting reactant?
F) What is the theoretical yield of
PbI2 in grams?
G) If the student obtained 0.583 grams
of PbI2 product after collecting it
by filtration and drying it, what
was the percent yield of PbI2
obtained?

Haven't had any lectures on this yet.

Can you please assist?

thank you

To solve these questions, we need to use the given reaction equation and the given masses of Pb(NO3)2 and KI. Here are the steps to calculate each part:

A) To find the number of moles of Pb(NO3)2 used:
- Convert the mass of Pb(NO3)2 to moles by dividing it by its molar mass.
- The molar mass of Pb(NO3)2 can be calculated by adding the atomic masses of lead (Pb), nitrogen (N), and oxygen (O).

B) To find the number of moles of KI used:
- Convert the mass of KI to moles by dividing it by its molar mass.
- The molar mass of KI can be calculated by adding the atomic masses of potassium (K) and iodine (I).

C) To find the number of moles of PbI2 formed based on the moles of Pb(NO3)2 used:
- Use the stoichiometric ratio between Pb(NO3)2 and PbI2 from the balanced equation.
- One mole of Pb(NO3)2 reacts with one mole of PbI2.

D) To find the number of moles of PbI2 formed based on the moles of KI used:
- Use the stoichiometric ratio between KI and PbI2 from the balanced equation.
- Two moles of KI react with one mole of PbI2.

E) To determine the limiting reactant:
- Compare the moles of PbI2 that can be formed from the moles of Pb(NO3)2 and KI.
- The reactant that produces fewer moles of PbI2 is the limiting reactant.

F) To find the theoretical yield of PbI2 in grams:
- Multiply the number of moles of PbI2 (based on the limiting reactant) by its molar mass.

G) To find the percent yield of PbI2 obtained:
- Divide the actual yield (0.583 grams) by the theoretical yield (calculated in the previous step) and multiply by 100.

If you provide the molar masses of Pb(NO3)2 and KI, I can help you with the calculations.

How much of this do you not know how to do? It's wasted time for me to go over steps you already know. And if you just want to check your answer, show your work and I'll check it.