100 ml of 0.1M NaCl solution is mixed with 100ml of 0.2M AgNo3 solution find:
(a)Mass of AgCl formed
(b) Limiting Agent
(c) mass of unreacted substance ( At. mass Na=23 , Ag=108 , Cl=35.5) ............ pls reply asap
Answers
NaCl + AgNO₃ -----------> NaNO₃ + AgCl
So, 1 mole of NaCl needs 1 mole of AgNO₃ to react completely
So, n AgNO₃ = M x V = 0.2 * 0.1 = 0.02 mole
n NaCl = M x V = 0.1 * 0.1 = 0.01 mole
As, AgNO₃ is present more than required, NaCl is our limiting agent
a) No. of moles of AgCl obtained = 0.01
Mass of AgCL formed = 0.01 * 143.5 [ mol.wt of AgCl ]
= 1.435 g of AgCl
b) AgNO₃ is present more than required, NaCl is our limiting agent.
c) Here AgNO₃ is present more than required (0.02 mole), NaCl is our limiting agent as it is present 0.01 mole less than AgNO₃. Thus unreacted or excess reagent is AgNO₃.
In our reaction 0.01 mole of NaCl require 0.01 mole out of 0.02 mole of AgNO₃ to form 0.01 mole of AgCl.
So, unreacted no. of moles of AgNO₃ = 0.01
Mass of AgNO₃ unreacted = 0.01 * 170
= 1.7 g of AgNO₃
Our balanced equation is:
NaCl + AgNO₃ -----------> NaNO₃ + AgCl
So, 1 mole of NaCl needs 1 mole of AgNO₃ to react completely
So, n AgNO₃ = M x V = 0.2 * 0.1 = 0.02 mole
n NaCl = M x V = 0.1 * 0.1 = 0.01 mole
As, AgNO₃ is present more than required, NaCl is our limiting agent
a) No. of moles of AgCl obtained = 0.01
Mass of AgCL formed = 0.01 * 143.5 [ mol.wt of AgCl ]
= 1.435 g of AgCl
b) AgNO₃ is present more than required, NaCl is our limiting agent.
c) Here AgNO₃ is present more than required (0.02 mole), NaCl is our limiting agent as it is present 0.01 mole less than AgNO₃. Thus unreacted or excess reagent is AgNO₃.
In our reaction 0.01 mole of NaCl require 0.01 mole out of 0.02 mole of AgNO₃ to form 0.01 mole of AgCl.
So, unreacted no. of moles of AgNO₃ = 0.01
Mass of AgNO₃ unreacted = 0.01 * 170
= 1.7 g of AgNO₃
Hope this helps.
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