if 34.19 grams of sodium chloride reacts with excess silver nitrate, how many grams of silver chloride can be produced?

Question

I need the balanced equation and type of reactio. Showing all work 

Answer 1

We start by identifying the stoichiometric equation for the reaction.

 

NaCl + AgNO₃ ---> AgCl + NaNO₃

 

This equation is already balanced, but always be sure to check!

 

Next we need to convert the mass of our NaCl into moles.

 

34.19 g NaCl * (1 mol NaCl / 58.44 g NaCl) = 0.585 mol NaCl

 

Since sodium chloride is the limiting reagent, this is the maximum number of moles available in the reaction. Convert to moles of silver chloride.

 

0.585 mol NaCl * (1 mol AgCl / 1 mol NaCl) = 0.585 mol AgCl   <--- This conversion factor comes from the stoichiometry

 

Then convert from moles of silver chloride to grams of silver chloride.

 

0.585 mol AgCl * (143.32 g AgCl / 1 mol AgCl) = 83.85 g AgCl

 

Therefore the total amount of silver chloride we can produce is 83.85 g.