Question

For the reactions shown below, we added 2.25 mL of 0.0460 M HCl to a test tube containing one of the two cations (Ag⁺ or Pb²⁺) and recovered 0.0148 g of precipitate.

AgNO₃(aq) + HCl(aq) → AgCl(s) + HNO₃(aq)
Pb(NO₃)₂(aq) + 2 HCl(aq) → PbCl₂(s) + 2 HNO₃(aq)

How much precipitate in moles would be recovered theoretically if the ion was Ag⁺? (Enter an unrounded value. Use at least one more digit than given.)


How much precipitate in moles would be recovered theoretically if the ion was Pb²⁺? (Enter an unrounded value. Use at least one more digit than given.)
mol

How much precipitate in grams would be recovered theoretically if the ion was Ag⁺?


How much precipitate in grams would be recovered theoretically if the ion was Pb²⁺?

Answer 1

For the reactions shown below, we added 2.25 mL of 0.0460 M HCl to a test tube containing one of the two cations (Ag⁺ or Pb²⁺) and recovered 0.0148 g of precipitate.

AgNO₃(aq) + HCl(aq) → AgCl(s) + HNO₃(aq)
Pb(NO₃)₂(aq) + 2 HCl(aq) → PbCl₂(s) + 2 HNO₃(aq)

How much precipitate in moles would be recovered theoretically if the ion was Ag⁺? (Enter an unrounded value. Use at least one more digit than given.)
From the balanced equation ,

1 mol HCl will produce 1 mol AgCl.

Mol of HCl in 2.25mL of 0.0460 M solution = 2.25/1000*0.0460 =0.0001035= 1.035*10^-4 mol HCl
This will produce 1.035*10^-4 mol AgCl ------answer

How much precipitate in moles would be recovered theoretically if the ion was Pb²⁺? (Enter an unrounded value. Use at least one more digit than given.)
mol

From the balanced equation 2 mol HCl will produce 1 mol PbCl2
From above you have added 1.035*10^-4 mol HCl
This will produce (1.035*10^-4) /2 =0.00005175= 5.175*10^-5 mol PbCl2
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How much precipitate in grams would be recovered theoretically if the ion was Ag⁺?

Molar mass AgCl = 143.32 g/mol

Mass of 1.035*10^-4 mol AgCl = 1.035*10^-4 * 143.32 =0.01483362= 0.0148 g AgCl precipitated

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How much precipitate in grams would be recovered theoretically if the ion was Pb²⁺?

Molar mass PbCl2 = 278.12 g/mol
Mass of 5.175*10^-5 mol PbCl2 = 5.175*10^-5 * 278.12 =0.01439271= 0.0143 g PbCl2 precipitated.