Question #d03c9

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Question #d03c9

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Answer 1 · 2015-01-31T16:37:19

$55.4 g$ $"55.4 g"$

Explanation:

Every time you have to solve problems in which you are asked to determine how much of something is needed to react with a given mass of something else, you must think moles and, more specifically, mole ratios.

Start with the balanced chemical equation

$2 K I (a q) + P b {(N O_{3})}_{2} (a q) \to P b I_{2} (s) + 2 K N O_{3} (a q)$ $2KI(aq) + Pb(NO_3)_2(aq) -> PbI_2(s) + 2KNO_3(aq)$

This is a double replacement reaction that leads to the formation of an insoluble precipitate, $P b I_{2}$ $PbI_2$ .

Notice the mole ratio between $K I$ $KI$ and $P b {(N O_{3})}_{2}$ $Pb(NO_3)_2$ : $2 moles$ $"2 moles"$ of the former must react with $1 mole$ $"1 mole"$ of the latter in order for the reaction to take place.

Use the molar masses of the compounds to go from grams to moles (for $K I$ $KI$ ) and from moles to grams (for $P b {(N O_{3})}_{2}$ $Pb(NO_3)_2$ ):

$55.2 g \cdot \frac{1 mole}{331.2 g} = 0.167 moles$ $55.2 cancel("g") * ("1 mole")/(331.2 cancel("g")) = "0.167 moles"$ $P b {(N O_{3})}_{2}$ $Pb(NO_3)_2$

This means that you need

$0.167 moles a P b {(N O_{3})}_{2} \cdot \frac{2 moles K I}{1 mole a P b {(N O_{3})}_{2}} = 0.334 moles a K I$ $0.167 cancel("moles"color(white)(a) Pb(NO_3)_2) * ("2 moles " KI)/(1 cancel("mole"color(white)(a) Pb(NO_3)_2)) = "0.334 moles"color(white)(a) KI$

The mass will be

$0.334 moles \cdot \frac{166 g}{1 mole} = 55.4 g a K I$ $0.334 cancel("moles") * ("166 g")/(1 cancel("mole")) = "55.4 g"color(white)(a) KI$

Here $166 g$ $"166 g"$ represents the mass of one mole of potassium iodide. This values comes from potassium iodide's molar mass, which is listed at approximately $166 g /mol$ $"166 g /mol"$ .

As a conclusion, always go to moles, it'll make your life easier, and remember to use mole ratios.

Here's a video of the reaction, if you're curios about what the precipitate looks like:

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Question #d03c9

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