Question

500 mL of 0.2 M NaCl sol, is added to 100 mL of 0.5 M AgNO3, solution resulting in the formation of white precipitate of AgCI. How many moles and how many grams of AgCl are formed ? Which is the limiting reagent?​

Answer 1

Answer:

Molarity of NaCl= 0.2 M

Volume of NaCl solution=500 ml=0.5 L

Number of moles of NaCl =Molarity×volume=0.2×0.5=0.1 moles____(ans 1)

Molarity of $AgNO_{3}$=0.5 M

Volume of $AgNO_{3}$ solution=100 ml =0.1 L

Number of moles of $AgNO_{3}$ =molarity×volume=0.5×0.1=0.05 moles____ (ans 2)

Number of moles of $AgNO_{3}$ <Number of moles of NaCl

=>$AgNO_{3}$ is the limiting reagent____(ans 3) ($\because$ limiting reagent gets completely consumed up in the reaction as it has lesser number of moles)

The proposed reaction is represented as $NaCl+AgNO_{3}=AgCl+NaNO_{3}+NaCl(0.05 moles)$

In the above reaction,

Number of moles of AgCl=1×number of moles of the limiting reagent=(1×0.05) moles=0.05 moles____(ans 4)

Gram Molecular weight of 1 mol of AgCl =(107.9+35.5)g=143.4 g

Gram molecular weight of 0.05 mol of AgCl =(0.05×143.4)g=7.17 g____(ans 5)