B----------- oxide is the correct answer
I believe the balanced chemical equation is:
C6H12O6 (aq) + 6O2(g)
------> 6CO2(g) + 6H2O(l)
First calculate the
moles of CO2 produced:
moles CO2 = 25.5 g
C6H12O6 * (1 mol C6H12O6 / 180.15 g) * (6 mol CO2 / 1 mol C6H12O6)
moles CO2 = 0.8493 mol
Using PV = nRT from
the ideal gas law:
<span>V = nRT / P</span>
V = 0.8493 mol *
0.08205746 L atm / mol K * (37 + 273.15 K) / 0.970 atm
<span>V = 22.28 L</span>
4
N
a
+
O
2
→
2
N
a
2
O
.
By the stoichiometry of this reaction if 5 mol natrium react, then 2.5 mol
N
a
2
O
should result.
Explanation:
The molecular mass of natrium oxide is
61.98
g
⋅
m
o
l
−
1
. If
5
m
o
l
natrium react, then
5
2
m
o
l
×
61.98
g
⋅
m
o
l
−
1
=
154.95
g
natrium oxide should result.
So what have I done here? First, I had a balanced chemical equation (this is the important step; is it balanced?). Then I used the stoichiometry to get the molar quantity of product, and converted this molar quantity to mass. If this is not clear, I am willing to have another go
