Answer:
2.07 mol O₂
Explanation:
First we need to write down the species present in the chemical equation, using the information given by the exercise:
However this equation <em>is not balanced</em>, so now we<u> balance it</u>:
Now we can use the stoichiometric ratio to <u>calculate the moles of oxygen </u>from the moles of sulfide dioxide:
- 1.38 molSO₂ *
= 2.07 mol O₂
Remember that in this case pressure is equal to 1.00 atm and temperature is equal to 273.15K. So,
P
V
=
n
R
T
→
n
=
P
V
R
T
=
1.00
a
t
m
⋅
7.0
L
0.082
a
t
m
⋅
L
m
o
l
⋅
K
⋅
273.15
K
=
0.31
Since we know hydrogen's molar mass (
2.0
g
m
o
l
), we can determine the mass
m
H
2
=
n
⋅
m
o
l
a
r
.
m
a
s
s
=
0.31
m
o
l
e
s
⋅
2.0
g
m
o
l
=
0.62
g
If indeed you are dealing with STP, remember that, under these conditions, 1 mole of any ideal gas occupies
22.4
L
. So,
n
=
V
V
m
o
l
a
r
=
7.0
L
22.4
L
=
0.31
moles
And, once again,
m
=
0.31
⋅
2.0
=
0.6
75.2 grams....????? Maybe
Answer:
1.08 M
Explanation:
From the question given above, the following data were obtained:
Initial volume (V₁) = 45 mL
Initial concentration (C₁) = 6 M
Final volume (V₂) = 250 mL
Final concentration (C₂) =?
The final concentration of the solution can be obtained by using the dilution formula as illustrated below:
C₁V₁ = C₂V₂
6 × 45 = C₂ × 250
270 = C₂ × 250
Divide both side by 250
C₂ = 270 / 250
C₂ = 1.08 M
Therefore, the final concentration of the solution is 1.08 M.
