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We wish to determine the moles of [tex]\(\text{PbI}_2\)[/tex] precipitated when 125 mL of 0.20 M KI reacts with excess [tex]\(\text{Pb}(\text{NO}_3)_2\)[/tex].

[tex]\[
2 \text{KI} (aq) + \text{Pb}(\text{NO}_3)_2 (aq) \rightarrow 2 \text{KNO}_3 (aq) + \text{PbI}_2 (s)
\][/tex]

In the previous step, you determined 0.025 mol KI react.

How many moles of [tex]\(\text{PbI}_2\)[/tex] form during the reaction?

Moles of [tex]\(\text{PbI}_2\)[/tex]:

Sagot :

GinnyAnswer
To find how many moles of [tex]\( PbI_2 \)[/tex] are formed during the reaction, we need to follow these steps:

1. Determine the stoichiometric relationship from the balanced chemical equation:

The balanced chemical equation is:
[tex]\[ 2 \, KI (aq) + Pb(NO_3)_2 (aq) \rightarrow 2 \, KNO_3 (aq) + PbI_2 (s) \][/tex]

According to the equation, 2 moles of [tex]\( KI \)[/tex] react with 1 mole of [tex]\( Pb(NO_3)_2 \)[/tex] to produce 1 mole of [tex]\( PbI_2 \)[/tex].

2. Given moles of [tex]\( KI \)[/tex]:

It was determined previously that there are 0.025 moles of [tex]\( KI \)[/tex] reacting.

3. Calculate moles of [tex]\( PbI_2 \)[/tex] produced:

Use the stoichiometric relationship from the balanced equation. For every 2 moles of [tex]\( KI \)[/tex], 1 mole of [tex]\( PbI_2 \)[/tex] is produced. Therefore:
[tex]\[ \text{moles of } PbI_2 = \frac{\text{moles of } KI}{2} \][/tex]
Given 0.025 moles of [tex]\( KI \)[/tex]:
[tex]\[ \text{moles of } PbI_2 = \frac{0.025 \, \text{moles}}{2} = 0.0125 \, \text{moles} \][/tex]

Thus, the moles of [tex]\( PbI_2 \)[/tex] formed during the reaction are [tex]\( 0.0125 \)[/tex] moles.
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