Astudent reacts 0.600 g of lead (ii) nitrate with 0.850 g of potassium iodidewrite and balance the chemical equation. how many miles of pb(no3)2 were used? how many miles of ki were used? how many miles of pbi2 would form based on the moles of pb(no3)2 used? how many miles of pbi2 would form based on the moles of ki used? which is the limiting reactant? what is the theoretical yeild of pbi2 in grams? if the student obtained 0.475 grams of pbi2 product after conducting it by filtration, what is the percent yeild of pbi2 obtained ?

Q1)
the reaction that takes place is 
lead nitrate reacting with potassium iodide to form lead iodide and potassium nitrate 
balanced chemical equation for the reaction is as follows
Pb(NO₃)₂ + 2KI > PbI₂  + 2KNO₃

Q2)
mass of lead nitrate present - 0.600 g 
number of moles = mass present / molar mass 
number of moles - 0.600 g / 331.2 g/mol = 0.00181 mol 

Q3)
mass of potassium iodide present - 0.850 g
number of moles = mass present / molar mass
number of moles of potassium iodide = 0.850 g / 166 g/mol = 0.00512 mol

Q4)
we have to calculate the number of moles of PbI₂ formed based on the number of moles of Pb(NO₃)₂ present assuming the whole amount of Pb(NO₃)₂ was used up 
stoichiometry of Pb(NO₃)₂ to PbI₂ is 1:1
number of Pb(NO₃)₂ moles reacted - 0.00181 mol
therefore number of PbI₂ moles formed - 0.00181 mol 

Q5)
next we have to calculate the number of moles of PbI₂ formed based on the amount of KI moles present , assuming all the moles of KI were used up in the reaction 
stoichiometry of KI to PbI₂ is 2:1
number of moles of KI reacted - 0.00512 mol
then number of moles of PbI₂ formed - 0.00512 x 2 = 0.0102 mol
0.0102 mol of PbI₂ is formed 

Q6)
limting reactant is the reactant that is fully consumed during the reaction. the amount of product formed depends on the amount of limiting reactant present

if lead nitrate is the limiting reactant 
if 1 mol of Pb(NO₃)₂ reacts with 2 mol of KI 
then 0.00181 mol of Pb(NO₃)₂ reacts with - 2 x 0.00181 mol of KI = 0.00362 mol 
but 0.00512 mol of KI is present and only 0.00362 mol are required 
therefore KI is in excess and Pb(NO₃)₂ is the limiting reactant 

Pb(NO₃)₂ is the limiting reactant 

Q7)
then the amount of PbI₂ formed depends on amount of Pb(NO₃)₂ present 
therefore number of moles of PbI₂ formed is based on number of Pb(NO₃)₂ moles present 
as calculated in Question number 4 - Q4
number of PbI₂ moles formed - 0.00181 mol 
mass of PbI₂ formed - 461 g/mol x 0.00181 mol = 0.834 g
mass of PbI₂ formed - 0.834 g

Q8) 
actual yield obtained  is not always equal to the theoretical yield . therefore we have to find the percent yield. This tells us the percentage of the theoretical yield that is actually obtained after the experiment
percent yield = actual yield / theoretical yield x 100 %
percent yield = 0.475 g / 0.834 g x 100 % = 57.0 %
percent yield of lead iodide is 57.0 %

The internal energy of the system we can term it as total potential energy and kinetic energy. energy flows out of the surrounding from the system then we can say that the internal energy of the final state is lower than its initials. the sign of  δesystem is negative.δesystem = efinal - einitial=-ve  efinal  ∠einitial.