Calculate the volume of hydrogen gas required for the formation of water in the following reaction 2H2 + O2 gives you 2H2O:
22.4 litre
11.2 litre
24 litre
44.8 litre
Answers
Explanation:
You'd need 6.55 L of oxygen at STP to form that much water.
STP conditions, which are defined as a pressure of 100 kPa and a temperature of 273.15 K, should immediately get you thinking about molar volume of a gas.
More specifically, about the fact that 1 mole of any ideal gas occupies exactly 22.7 L at STP.
This means that all you really need to figure out is the number of moles of oxygen that took part in the reaction. If you know how many moles of oxygen are needed, you can use the molar volume of a gas at STP to calculate the volume of oxygen.
Use water's molar mass to determine how many moles of water would the reaction produce
10.5
g
⋅
1 mole water
18.02
g
=
0.577 moles
H
2
O
Now take a look at the balanced chemical equation
2
H
2
(
g
)
+
O
2
(
g
)
→
2
H
2
O
(
g
)
Notice the
1
:
2
mole ratio that exists between oxygen and water. This means that 1 mole of oxygen will produce 2 moles of water.
Calculate the number of moles of oxygen by
0.577
moles
H
2
O
⋅
1 mole
O
2
2
moles
H
2
O
=
0.2885 moles
O
2
So, if 1 mole occupies 22.7 L at STP, then
0.2885
moles
⋅
22.7 L
1
mole
=
6.548 L
Rounded to three sig figs, the answer will be
V
O
2
=
6.55 L
SIDE NOTE More often than not, you will be required to use the old definition of STP, which implies a pressure of 1 atm and a temperature of 273.15 K. Under these conditions, 1 mole of any ideal gas occupies 22.4 L.
The current definition of STP implies a molar volume of 22.7 L, but if your instructor or teacher wants you to use the old value, simply redo the final calculation using 22.4 instead of 22.7 L
Answer:
44.8 L
Explanation: