How many minutes does it take to form 10L of O2 measured at 92.2 kPa and 28°c from water if a current of 1.3 A passes through the electolytic cell?
a) how many minutes?
b)What mass in grams of h2 forms?
To calculate how many minutes it takes to form 10L of O2, we need to determine the amount of time it takes for a certain amount of charge to pass through the electrolytic cell. Here's how we can solve this:
a) First, we need to determine the total charge passed through the cell. The charge (Q) can be calculated using the following formula:
Q = I * t
Where:
Q = charge
I = current (in Amperes)
t = time (in seconds)
Since the question asks for the time in minutes, we need to convert it. There are 60 seconds in a minute, so we divide the time by 60.
Now, let's calculate the charge:
Q = (1.3 A) * (t/60)
b) The formation of O2 in an electrolytic cell occurs through the reduction of water. The balanced chemical equation for the reaction is:
2 H2O(l) -> 2 H2(g) + O2(g)
From this equation, we can see that for every 2 moles of H2O, we get 1 mole of O2 gas. Since we know the volume of O2 (10 L), we can use the ideal gas law to calculate the amount of O2 in moles.
PV = nRT
Where:
P = pressure (92.2 kPa)
V = volume (10 L)
n = moles
R = gas constant (0.0821 L·atm/(mol·K))
T = temperature (in Kelvins)
First, let's convert the pressure from kPa to atm:
1 atm = 101.325 kPa
92.2 kPa * (1 atm / 101.325 kPa) = 0.9093 atm
Now, we can solve for n:
n = (P * V) / (R * T)
Since we are given the temperature in degrees Celsius, we need to convert it to Kelvins:
T(K) = T(°C) + 273.15 = 28 + 273.15 = 301.15 K
n = (0.9093 atm * 10 L) / (0.0821 L·atm/(mol·K) * 301.15 K)
Now we can calculate the mass of H2 formed. Since we know the molar mass of H2 is 2 g/mol, we can convert moles to grams:
m(g) = n(mol) * molar mass(g/mol)
m = (n * 2 g/mol)
Now we have successfully derived the answers to both parts (a) and (b) of the question.