Answer :
Answer
the limiting reagent is NaCl
The mass of AgCl to be produced = 34.8 g
Explanation
Given:
Balanced chemical equation: 1AgNO3+1NaCl=1AgCl+1NaNO3
mass of AgCl = ?
AgNO3 = 53.42 g
NaCl = 14.19 g
Solution
Step 1: Calculate the number of moles of reactants, to find the limiting reactant/reagent.
For AgNO3:
[tex]53.42g\text{ AgNO}_3\text{ x }\frac{1\text{ mol AgNO}_3}{169,87g}\text{ x }\frac{1\text{ AgCl}}{1\text{ AgNO}_3}\text{ = 0.314 mol AgCl}[/tex]For NaCl
[tex]14.19g\text{ NaCl x }\frac{1\text{ mol NaCl}}{58,44g}\text{ x }\frac{1\text{ AgCl}}{1\text{ NaCl}}\text{ = 0.243 mol AgCl}[/tex]Therefore the limiting reagent is NaCl
Step 2: Calculate the mass of AgCl
[tex]0.243\text{ mol AgCl x }\frac{143,32g\text{ AgCl}}{1mol\text{ AgCl}}\text{ = 34.8 g AgCl}[/tex]