Calculate the volume of oxygen at 10 degrees Celsius and 1.6 atm pressure that can be produced by decomposing 128g of hydrogen peroxide
To calculate the volume of oxygen produced by decomposing hydrogen peroxide, you need to use the ideal gas law equation:
PV = nRT
Where:
P = pressure (in atm)
V = volume (in liters)
n = number of moles of gas
R = ideal gas constant (0.0821 L·atm/mol·K)
T = temperature (in Kelvin)
We'll first convert the given Celsius temperature to Kelvin:
T(K) = T(°C) + 273.15
So, T(K) = 10 + 273.15 = 283.15 K
Next, we need to calculate the number of moles of oxygen produced from the given mass of hydrogen peroxide. Hydrogen peroxide (H2O2) decomposes into water (H2O) and oxygen (O2) according to the following balanced equation:
2 H2O2 -> 2 H2O + O2
The molar mass of hydrogen peroxide is 34.02 g/mol, so the number of moles can be calculated as follows:
n = mass / molar mass
n = 128 g / 34.02 g/mol ≈ 3.764 mol
Now, we can substitute the values into the ideal gas law equation:
PV = nRT
V = (nRT) / P
V = (3.764 mol)(0.0821 L·atm/mol·K)(283.15 K) / 1.6 atm
V ≈ 13.01 L
Therefore, the volume of oxygen that can be produced by decomposing 128g of hydrogen peroxide at 10 degrees Celsius and 1.6 atm pressure is approximately 13.01 liters.