What is the purpose of using excess AgNO3?

Answer: To ensure complete precipitation of all Cl- as AgCl Excess Ag+ drives the precipitation of Cl- to completion (very small Ksp of AgCl) so that the gravimetric analysis is quantitative.

Which is the limiting reagent?

Answer: Both are stoichiometric — neither is limiting. Need 2 mol H₂ per 1 mol O₂. We have exactly 2.0 mol H₂ and 1.0 mol O₂ — perfect stoichiometric ratio, no limiting reagent.

The percent yield is:

Answer: 80.0% %yield = (actual/theoretical)100 = (20.0/25.0)100 = 80.0%.

AP Chemistry · Topic 4.5

Stoichiometry Practice

Part of Chemical Reactions.(SPQ-4.A)

Practice questions

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Sample questions

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Sample 1difficulty 1/5
$To determine the chloride content of an unknown soluble chloride salt, a student dissolves 0.4250 g of the salt in water and adds excess 0.10 M AgNO3. A white precipitate forms, which is filtered onto pre-weighed filter paper, washed with cold water, and dried to constant mass. Mass of dried precipitate = 0.8610 g.$
$What is the purpose of using excess AgNO3?$
- A
  $To ensure complete precipitation of all Cl- as AgCl$
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- B
  $To form a more soluble silver complex$
- C
  $To decrease the molar mass of the precipitate$
- D
  $To buffer the solution at a fixed pH$
Why
$Excess Ag+ drives the precipitation of Cl- to completion (very small Ksp of AgCl) so that the gravimetric analysis is quantitative.$
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Sample 2difficulty 2/5
$For 2 H₂ + O₂ \to 2 H₂O, 4.0 g H₂ and 32.0 g O₂ are mixed.$
$Which is the limiting reagent?$
- A
  $O₂$
- B
  $Both are stoichiometric — neither is limiting.$
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- C
  $Cannot determine.$
- D
  $H₂$
Why
$Need 2 mol H₂ per 1 mol O₂. We have exactly 2.0 mol H₂ and 1.0 mol O₂ — perfect stoichiometric ratio, no limiting reagent.$
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Sample 3difficulty 2/5
$The percent yield is:$
- A
  $80.0%$
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- B
  $5.0%$
- C
  $125%$
- D
  $20.0%$
Why
$%yield = (actual/theoretical)*100 = (20.0/25.0)*100 = 80.0%.$
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Sample 4difficulty 2/5
$For 2 H₂ + O₂ \to 2 H₂O: how many moles of O₂ react with 4 mol H₂?$
- A
  $2 mol$
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- B
  $4 mol$
- C
  $8 mol$
- D
  $1 mol$
Why
$Mole ratio H₂ : O₂ = 2 : 1. So 4 mol H₂ requires 2 mol O₂.$
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Sample 5difficulty 2/5
$A student investigates the limiting reactant in the reaction Pb(NO3)2 + 2 KI -> PbI2 + 2 KNO3 by mixing constant total volume (50.0 mL) of 0.10 M Pb(NO3)2 and 0.10 M KI in five different ratios (Job's plot) and measuring the dry mass of yellow PbI2 precipitate. Maximum yield occurs at V(Pb)=16.7 mL, V(KI)=33.3 mL.$
$What does the position of the maximum confirm?$
- A
  $Pb^2+ and I- combine 1:1$
- B
  $PbI2 is soluble$
- C
  $KI is the spectator$
- D
  $The stoichiometric ratio of Pb^2+ : I- in the precipitate is 1:2$
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Why
$A maximum yield at 1:2 mol ratio of Pb^2+ : I- (16.7:33.3 mL of equimolar solutions) is consistent with the formula PbI2.$
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