Sodium and chlorine gas react to produce sodium chloride. If $"234 g"$ of sodium chloride is produced, how much sodium reacted?

Md Ariful Islam

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Md Ariful Islam

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You know that

$2"Na"_ ((s)) + "Cl"_ (2(g)) -> 2"NaCl"_ ((s))$

and that the reaction produced $"234 g"$ of sodium chloride. Convert this to moles by using the compound's molar mass

$234 color(red)(cancel(color(black)("g"))) * "1 mole NaCl"/(58.44 color(red)(cancel(color(black)("g")))) = "4.004 moles NaCl"$

Now, you know that the sample of sodium metal reacted completely, which implies that it was completely consumed by the reaction.

In other words, you don't have to worry about the sample of chlorine gas because the fact that sample of sodium metal was completely consumed lets you know that the chlorine gas is not a .

This means that the reaction consumed

$4.004 color(red)(cancel(color(black)("moles NaCl"))) * "2 moles Na"/(2color(red)(cancel(color(black)("moles NaCl")))) = "4.004 moles Na"$

Convert this to grams by using the element's molar mass

$4.004 color(red)(cancel(color(black)("moles Na"))) * "22.99 g"/(1color(red)(cancel(color(black)("mole Na")))) = color(darkgreen)(ul(color(black)("92.1 g")))$

The answer is rounded to three .

SIDE NOTE You can show that chlorine gas is not the limiting reagent by calculating the number of moles present in the sample

$0.142 color(red)(cancel(color(black)("kg"))) * "1 mole Cl"_2/(35.453color(red)(cancel(color(black)("kg")))) = "4.005 moles Cl"_2$

In order to produce $4.004$ moles of sodium chloride, you only need

$4.004 color(red)(cancel(color(black)("moles NaCl"))) * "1 mole Cl"_2/(2color(red)(cancel(color(black)("moles NaCl")))) = "2.002 moles Cl"_2$

Since you have more chlorine gas than you need, chlorine is in excess.

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