what mass of silver iodide can be made by the reaction of 10g of silver nitrate with 10g of sodium

Question

I really am stuck on this problem and need help!

Answer 1

Hi Chris,

I'm going to assume that second part is 10g of sodium idide. Here goes..

  1 AgNO3  +  1 NaI   --> 1  AgI  + 1 NaNO3

 

 

10g AgNO3 x 1 mol AgNO3  X 1 mol AgI      x     235 gAgI

_____________________________________________         =  13.8 g AgI

                  170g AgNO3    x  1 mol AgNO3 x    1 mol AgI

 

 

 

10g NaI    x   1 mol NaI    x     1 mol  AgI    x  235g  AgI

_________________________________________________  =  15.7 g AgI

                    150g NaI     x     1 mol NaI    x  1 mol AgI    

 

So after making 13.8g of AgI, we have used all 10g of AgNO3 and the reaction stops.  Therefore,  13.8g of AgI is the max that can be made with 10g of each reactant.  AgNO3 is called the limiting reactant.  There will be excess Nai left at the end of the reaction.  Hope that helped!