A certain quantity of electricity deposits 0.54grams of silver nitrate solution. What volume of H2 will be liberated by the same quantity of electricity at 27 ° C and 750 mm of Hg pressure
To determine the volume of H2 that will be liberated by the given quantity of electricity, we need to apply Faraday's laws of electrolysis.
First, we need to find the number of moles of silver nitrate (AgNO3) that will be deposited by the given quantity of electricity. We can then use this to find the number of moles of H2 produced, assuming a balanced chemical equation.
Step 1: Find the number of moles of AgNO3 deposited.
To do this, we need to know the molar mass of AgNO3, which is the sum of the atomic masses of its constituent elements:
Ag: 107.87 g/mol
N: 14.01 g/mol
O: 16.00 g/mol (x3)
Summing them up: 107.87 + 14.01 + (16.00 x 3) = 169.87 g/mol
Given mass of silver nitrate deposited = 0.54 grams
Number of moles of AgNO3 = mass / molar mass = 0.54 g / 169.87 g/mol ≈ 0.00318 mol
Step 2: Determine the number of moles of H2 produced.
The balanced chemical equation for the electrolysis of water is:
2 H2O(l) → 2 H2(g) + O2(g)
From the equation, we see that for every 2 moles of water (H2O) that react, we obtain 2 moles of H2 gas. Therefore, the moles of H2 produced will be the same as the moles of water consumed.
Moles of H2 produced = Moles of water consumed = 0.00318 mol
Step 3: Calculate the volume of H2 gas liberated using the ideal gas law.
The ideal gas law is given by the equation: PV = nRT, where:
P = pressure (750 mmHg)
V = volume (unknown; what we are trying to find)
n = moles of gas (0.00318 mol)
R = ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature in Kelvin (27 °C + 273.15 = 300.15 K)
Rearranging the equation to solve for V:
V = (nRT) / P
Plugging in the values:
V = (0.00318 mol * 0.0821 L·atm/(mol·K) * 300.15 K) / 750 mmHg
Note: To ensure consistent units, we need to convert mmHg to atm by dividing by 760:
V = (0.00318 mol * 0.0821 L·atm/(mol·K) * 300.15 K) / (750 mmHg / 760 mmHg/atm)
Calculating the volume, we get:
V ≈ 0.01024 L = 10.24 mL
Therefore, the volume of H2 gas liberated by the given quantity of electricity at 27 °C and 750 mmHg pressure is approximately 10.24 mL.
To determine the volume of hydrogen gas liberated by a certain quantity of electricity, we need to use the ideal gas law equation:
PV = nRT
Where:
P = Pressure (in atm)
V = Volume (in liters)
n = Number of moles
R = Ideal gas constant (0.0821 L.atm/mol.K)
T = Temperature (in Kelvin)
First, let's convert the given pressure and temperature to the appropriate units:
Given:
Temperature (T) = 27 °C
Pressure (P) = 750 mmHg
Converting the temperature to Kelvin:
T = 27 + 273 = 300 K
Converting the pressure to atm:
P = 750 mmHg * (1 atm/760 mmHg) = 0.987 atm (rounded to 3 decimal places)
Now, let's calculate the number of moles of hydrogen gas (H2) using the given quantity of electricity that deposits 0.54 grams of silver nitrate solution. We need to consider the balanced chemical equation for the electrolysis of water:
2H2O(l) -> 2H2(g) + O2(g)
1 mole of H2 is liberated for every 2 moles of silver nitrate (AgNO3) deposited.
Given mass of silver nitrate (AgNO3) = 0.54 grams
Molar mass of AgNO3 = 169.87 g/mol
Number of moles of AgNO3 = mass / molar mass
Number of moles of AgNO3 = 0.54 g / 169.87 g/mol ≈ 0.00318 mol
Since 1 mole of AgNO3 produces 2 moles of H2, the number of moles of H2 gas = 2 * 0.00318 mol ≈ 0.00636 mol
Now, we can substitute the values into the ideal gas law equation:
PV = nRT
V = (nRT) / P
V = (0.00636 mol * 0.0821 L.atm/mol.K * 300 K) / 0.987 atm
V ≈ 0.1545 L (rounded to 4 decimal places)
Therefore, the volume of H2 gas liberated by the same quantity of electricity under the given conditions is approximately 0.1545 liters.
A poorly worded question. No amount of electricity will deposit AgNO3 solution. It will deposit Ag (metal) but not AgNO3 solution.
96,485 Coulombs will deposit 107.9 g Ag. How many C were involved? That's
96,485 x (0.54/107.9) = ?
That will liberate 1 g H2. Convert 1 g H2 to mols and use PV = nRT to convert to L at the conditions listed.