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
B
. They are large and occur at shallow depths near the where the plates diverge.
Answer:
%N = 25.94%
%O = 74.06%
Explanation:
Step 1: Calculate the mass of nitrogen in 1 mole of N₂O₅
We will multiply the molar mass of N by the number of N atoms in the formula of N₂O₅.
m(N): 2 × 14.01 g = 28.02 g
Step 2: Calculate the mass of oxygen in 1 mole of N₂O₅
We will multiply the molar mass of O by the number of O atoms in the formula of N₂O₅.
m(O): 5 × 16.00 g = 80.00 g
Step 3: Calculate the mass of 1 mole of N₂O₅
We will sum the masses of N and O.
m(N₂O₅) = m(N) + m(O) = 28.02 g + 80.00 g = 108.02 g
Step 4: Calculate the percent composition of N₂O₅
We will use the following expression.
%Element = m(Element)/m(Compound) × 100%
%N = m(N)/m(N₂O₅) × 100% = 28.02 g/108.02 g × 100% = 25.94%
%O = m(O)/m(N₂O₅) × 100% = 80.00 g/108.02 g × 100% = 74.06%
The answer is an igneous rock.
Hope this helps!!
