To find the volume of O2 liberated, we need to use the stoichiometry of the reaction.
From the balanced equation, we see that 1 mole of NaClO gives 1 mole of O2. We need to find the amount of NaClO in moles first.
To find moles of NaClO:
Moles = Mass / Molar Mass
Given:
Volume of bleach = 15 ml
Density of bleach = 1.12 g/ml
Percent composition of NaClO = 5.85%
First, we can calculate the mass of bleach:
Mass of bleach = Volume of bleach x Density of bleach
Mass of bleach = 15 ml x 1.12 g/ml
Next, we can find the mass of NaClO using the percent composition:
Mass of NaClO = Mass of bleach x (% NaClO / 100)
Mass of NaClO = (15 ml x 1.12 g/ml) x (5.85 / 100)
Now, we can find the moles of NaClO using the molar mass of NaClO:
Molar mass of NaClO = (23 + 35.5 + 16) g/mol
Moles of NaClO = Mass of NaClO / Molar mass of NaClO
Once we have the moles of NaClO, we know that it will produce an equal number of moles of O2. Then we can use the ideal gas law to calculate the volume of O2 in liters:
PV = nRT
Given:
Temperature (T) = 22°C = 22 + 273 = 295 K
Pressure (P) = 755 mmHg
R = 0.0821 L·atm/(mol·K) (gas constant)
Rearranging the equation:
V = nRT / P
Substitute the values in and solve for V.
Finally, convert the volume from liters to mL by multiplying by 1000.
Note: Make sure to convert the units used in the calculations to be consistent (e.g., grams, moles, atmosphere).
Now you have all the steps to solve the problem. Plug in the values you have obtained in each step and perform the necessary calculations to find the volume of O2 liberated.