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Ierofanga [76]
3 years ago
15

Potassium iodide reacts with lead(II) nitrate in this precipitation reaction: 2 KI(aq) + Pb(NO3)2(aq) → 2 KNO3(aq) + PbI2(s) Wha

t minimum volume of 0.200 M potassium iodide solution is required to completely precipitate all of the lead in 155.0 mL of a 0.112 M lead(II) nitrate solution?a. 174 mL
b. 86.8 mL
c. 189 mL
d. 94.6 mL
Chemistry
1 answer:
andrew11 [14]3 years ago
5 0

Answer:

a. 174 mL

Explanation:

Let's consider the following reaction.

2 KI(aq) + Pb(NO₃)₂(aq) → 2 KNO₃(aq) + PbI₂(s)

We have 155.0 mL of a 0.112 M lead(II) nitrate solution. The moles of Pb(NO₃)₂ are:

0.1550 L × 0.112 mol/L = 0.0174 mol

The molar ratio of KI to Pb(NO₃)₂ is 2:1. The moles of KI are:

2 × 0.0174 mol = 0.0348 mol

The volume of a 0.200 M KI solution that contains 0.0348 moles is:

0.0348 mol × (1 L / 0.200 mol) = 0.174 L = 174 mL

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\boxed{\text{20.2 mL}}

Explanation:

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\text{Moles of NH}_{3} = \text{253.5 g NH}_{3}\times \dfrac{\text{1 mol NH}_{3}}{\text{17.03 g NH}_{3}}= \text{14.89 mol NH}_{3}

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c₂ = 0.500 mol·L⁻¹; V₂ = 600 mL

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\begin{array}{rcl}14.89V_{1} & = & 0.500 \times 600\\14.89V_{1} & = & 300\\V_{1} & = & \text{20.2 mL}\\\end{array}\\\text{You will need $\boxed{\textbf{20.2 mL}}$ of the stock solution.}

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