The weight of AgCl precipitated when a solution containing 5.85 g of NaCl is added to a solution containing 3.4 g of AgNO3 is
  • a)
    28 g
  • b)
    9.25 g
  • c)
    2.870 g
  • d)
    58 g
Correct answer is option 'C'. Can you explain this answer?

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Answers

Universal Academy
Jan 19, 2022
AgNO3 + NaCl → AgCl + NaNO3
No. of moles of AgNO3 = 3.4/170 = 0.02
No, of moles of NaCl = 5.85/58.5 = 0.1
Limiting reagent = AgNO3
1 mole of AgNO3 produces 1 mole of AgCl
0.02 mole of AgNO3 will produce 0.02 mole of AgCl
Weight of AgCl produced = 0.02 x 143.5 = 2.870 g.

AgNO3 + NaCl → AgCl + NaNO3No. of moles of AgNO3 = 3.4/170 = 0.02No, of moles of NaCl = 5.85/58.5 = 0.1Limiting reagent = AgNO31 mole of AgNO3 produces 1 mole of AgCl0.02 mole of AgNO3 will produce 0.02 mole of AgClWeight of AgCl produced = 0.02 x 143.5 = 2.870 g.
AgNO3 + NaCl → AgCl + NaNO3No. of moles of AgNO3 = 3.4/170 = 0.02No, of moles of NaCl = 5.85/58.5 = 0.1Limiting reagent = AgNO31 mole of AgNO3 produces 1 mole of AgCl0.02 mole of AgNO3 will produce 0.02 mole of AgClWeight of AgCl produced = 0.02 x 143.5 = 2.870 g.