Answer:
Explanation:
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
g
⋅
1 mole NaCl
58.44
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 limiting reagent.
This means that the reaction consumed
4.004
moles NaCl
⋅
2 moles Na
2
moles NaCl
=
4.004 moles Na
Convert this to grams by using the element's molar mass
4.004
moles Na
⋅
22.99 g
1
mole Na
=
92.1 g
−−−−−
The answer is rounded to three sig figs.
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
kg
⋅
1 mole Cl
2
35.453
kg
=
4.005 moles Cl
2
In order to produce
4.004
moles of sodium chloride, you only need
4.004
moles NaCl
⋅
1 mole Cl
2
2
moles NaCl
=
2.002 moles Cl
2
Since you have more chlorine gas than you need, chlorine is in excess.vExplanation:
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
g
⋅
1 mole NaCl
58.44
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 limiting reagent.
This means that the reaction consumed
4.004
moles NaCl
⋅
2 moles Na
2
moles NaCl
=
4.004 moles Na
Convert this to grams by using the element's molar mass
4.004
moles Na
⋅
22.99 g
1
mole Na
=
92.1 g
−−−−−
The answer is rounded to three sig figs.
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
kg
⋅
1 mole Cl
2
35.453
kg
=
4.005 moles Cl
2
In order to produce
4.004
moles of sodium chloride, you only need
4.004
moles NaCl
⋅
1 mole Cl
2
2
moles NaCl
=
2.002 moles Cl
2
Since you have more chlorine gas than you need, chlorine is in excess.
Explanation:
