Answer to Question #305444 in General Chemistry for Mcquay

Question

Pb(NO₃)₂ + 2 NaI à PbI₂ + 2 NaNO₃                                  

1)     How many moles of lead (II) nitrate are needed to react completely with 10 moles of sodium iodide?

 

 

2)   How many moles of lead (II) iodide were produced when 5 moles of lead (II)nitrate were used in the reaction?

 

 

3)   If 100 grams of lead (II) nitrate were used in the reaction, how many grams of sodium iodide were used for a complete reaction?

 

4)   If 100 grams of lead (II) nitrate and 100 grams of sodium iodide were initially used in the reaction, which is the limiting reactant? Which is the excess reactant?

 

 

5)   Find the amount excess in grams in #4

 

 

6)   Suppose in the reaction in which 100 grams of lead(II) nitrate were all used up and only 100 grams of lead(II)iodide were obtained in the experiment, what is the percent yield?

 

 

7)   How many grams of the product were formed in no. 6 if the percent yield is 80%?

Answer 1

1)Answer=0.0362

:Because the mole ratio between Pb(NO₃)₂ and Pbl₂ is 1:1

Therefore,if 0.0362moles of Pb(NO₃)₂ reacts,,0.0362 moles of Pbl₂ forms.

2)0.0181 moles.

3)100 grams

4)Lead (Ii) nitrate is the limiting reagent.

Sodium iodide is the excess reagent