Question

500 ml of 0.200M NaCl solutions added to 100 ml of 0.500MAgNO3 solution resulting in the formation of white ppt of insoluble AgCl. What is the limiting reagent? How many moles how many gms of AgCl are formed ? Molar mass of Agcl = 143.5g per mol.

Answer 1

Answer: moles- 0.05 and mass - 7.175 g

Explanation:

no. of moles (n) = M x V( L)

no. of moles of NaCl = .200 x 500/1000

n of NaCl = 200/1000 x 1/2

n of NaCl = 1/10

no. of moles of AgNO3 = M x V

n of AgNO3 = 500/1000 x 100/1000

n of AgNO3 = 1/20

Now,

We can see that AgNO3 has 1/20 mole and NaCl has 1/10 moles therefore it is clear that 1/20 mole is less than 1/10 mole hence AgNO3 is limiting reagent

NaCl + AgNO3 == AgCl + NaNO3

0.1 mole 0.05 mol. 0.05mol

moles of AgCl = Mass / MolarMass

0.05 = mass/143.5

0.05 * 143.5 = Mass

Mass = 7.175 gram