Answer:
Explanation:
Sodium bicarbonate,
NaHCO
3
, will decompose to form sodium carbonate,
Na
2
CO
3
, water, and carbon dioxide,
CO
2
2
NaHCO
3(s]
→
Na
2
CO
3(s]
+
CO
2(g]
+
H
2
O
(g]
Notice that you have a
2
:
1
mole ratio between sodium bicarbonate and sodium carbonate. This means that the reaction will produce half as many moles of the latter than whatever number of moles of the former underwent decomposition.
Use sodium carbonate's molar amss to determine how many moles you'd get in that sample
0.685
g
⋅
1 mole NaHCO
3
84.007
g
=
0.008154 moles NaHCO
3
Now, if the reaction were to have a
100
%
yield, it would produce
0.008154
moles NaHCO
3
⋅
1 mole Na
2
CO
3
2
moles NaHCO
3
=
0.004077 moles Na
2
CO
3
Use the molar mass of sodium carbonate to determine how many grams would contain this many moles
0.004077
moles
⋅
105.99 g
1
mole
=
0.4321 g Na
2
CO
3Sodium bicarbonate,
NaHCO
3
, will decompose to form sodium carbonate,
Na
2
CO
3
, water, and carbon dioxide,
CO
2
2
NaHCO
3(s]
→
Na
2
CO
3(s]
+
CO
2(g]
+
H
2
O
(g]
Notice that you have a
2
:
1
mole ratio between sodium bicarbonate and sodium carbonate. This means that the reaction will produce half as many moles of the latter than whatever number of moles of the former underwent decomposition.
Use sodium carbonate's molar amss to determine how many moles you'd get in that sample
0.685
g
⋅
1 mole NaHCO
3
84.007
g
=
0.008154 moles NaHCO
3
Now, if the reaction were to have a
100
%
yield, it would produce
0.008154
moles NaHCO
3
⋅
1 mole Na
2
CO
3
2
moles NaHCO
3
=
0.004077 moles Na
2
CO
3
Use the molar mass of sodium carbonate to determine how many grams would contain this many moles
0.004077
moles
⋅
105.99 g
1
mole
=
0.4321 g Na
2
CO
3
