calculate the volume of hydrogen released when zinc reacted with 294 grams of sulfuric acid
Answers
Explanation:
Explanation:
Zinc metal will react with dilute sulfuric acid to produce zinc sulfate,
ZnSO
4
, and hydrogen gas,
H
2
, which will bubble out of solution.
The balanced chemical equation for this single replacement reaction looks like this
Zn
(s]
+
H
2
SO
4(aq]
→
ZnSO
4(aq]
+
H
2(g]
↑
⏐
Notice that you have a
1
:
1
mole ratio between zinc metal and hydrogen gas. This tells you that the reaction will always produce the same number of moles of hydrogen gas as you have moles of zinc metal that take part in the reaction.
In your case, you know that the sulfuric acid is in excess, which means that zinc metal acts as a limiting reagent, i.e. it will be completely consumed by the reaction.
Use zinc metal's molar mass to determine how many moles you have in that
4.33-g
sample
4.33
g
⋅
1 mole Zn
65.38
g
=
0.06623 moles Zn
This means that the reaction produced
0.06623
moles of hydrogen gas.
Now, in order to determine the volume this much hydrogen gas occupies under those conditions for pressure and temperature, you must use the ideal gas law equation, which looks like this
P
V
=
n
R
T
, where
P
- the pressure of the gas
V
- the volume it occupies
n
- the number of moles of gas
R
- the universal gas constant, usually given as
0.0821
atm
⋅
L
mol
⋅
K
T
- the temperature of the gas, always expressed in Kelvin!
It's important to notice here that the units given to you for pressure and temperature do not match those used in the expression of the universal gas constant.
This means that you have to convert these units from torr to atm, and from degrees Celsius to Kelvin, respectively.
So, rearrange the ideal gas law equation to solve for
V
, the volume of the gas
V
=
n
R
T
P
V
=
0.06623
moles
⋅
0.0821
atm
⋅
L
moles
⋅
K
⋅
(
273.15
+
38
)
K
760
760
atm
V
=
1.692 L
Rounded to two sig figs, the number of sig figs you have for the pressure and temperature of the gas, the answer will be
V
=
1.7 L
Given:
Mass of sulfuric acid = 294 gm
To Find:
The volume of hydrogen released when zinc reacts with the given mass of sulfuric acid.
Calculation:
- The reaction of zinc with sulfuric acid is given as:
Zn + H2SO4 → ZnSO4 + H2
- According to the given reaction:
Hydrogen produced when zinc reacts with 98 gm of Sulfuric acid = 1 mole
⇒ Hydrogen produced when zinc reacts with 294 gm of Sulfuric acid = (294/98) = 3 moles
- Volume of 3 mole hydrogen gas, V = 3 × 22.4
⇒ V = 67.2 L
- So, the volume of hydrogen released when zinc reacts with 294 gm of sulfuric acid is 67.2 L.