Water gas (CO + H2) consisting of equal volumes of CO and H2 is produc...
Calculation of Volume of Water Gas Produced
Given:
- Water gas (CO + H2) is produced when water vapor is passed over red hot coal.
- The volume of water gas produced is equal to the sum of the volumes of CO and H2 produced.
- The volumes of CO and H2 are equal.
- 1 mole of any gas occupies 22.4L under standard conditions.
Step 1: Calculate the moles of CO and H2 produced
Since the volumes of CO and H2 are equal, we can assume that the volume of each gas is x.
The balanced equation for the reaction is:
C + H2O → CO + H2
From the equation, we can see that 1 mole of C reacts with 1 mole of H2O to produce 1 mole of CO and 1 mole of H2.
Given that 3 kg of coal is treated, we need to convert the mass of coal to moles of C:
Molar mass of C = 12 g/mol
Number of moles of C = (3 kg) / (12 g/mol) = 250 mol
Therefore, the number of moles of CO and H2 produced is also 250 mol.
Step 2: Calculate the volume of water gas produced
Since the volume of each gas is x, the total volume of water gas produced is 2x (CO + H2).
According to the ideal gas law, the volume of a gas is directly proportional to the number of moles of the gas:
V = nRT/P
Under standard conditions, the pressure (P) is 1 atm and the temperature (T) is 273 K.
Substituting the values into the equation:
V = (2x)(22.4 L/mol)(273 K) / (1 atm)
V = 12121.6x L
Step 3: Determine the range of the volume
Since the volume of water gas produced is directly proportional to x, we need to find the range of x that corresponds to the given range of the volume of water gas produced.
Let's assume that the volume of water gas produced is within the range of 11195 L to 11205 L.
Substituting the values into the equation:
11195 L ≤ 12121.6x L ≤ 11205 L
Dividing all sides of the inequality by 12121.6:
0.9227 ≤ x ≤ 0.9235
Therefore, the range of x is approximately 0.9227 to 0.9235.
Step 4: Calculate the volume of water gas produced
To find the volume of water gas produced, we need to calculate the volume of each gas (CO and H2) and then sum them.
Volume of CO = x = 0.9227 L
Volume of H2 = x = 0.9227 L
Total volume of water gas produced = 2x = 2(0.9227 L) = 1.8454 L
Therefore, the volume of water gas produced when 3 kg of coal is treated with water vapor is approximately 1.8454 L, which falls within the given range of 11195 L to 11205 L.