Oxygen gas can be generated by heating KClO3 in the presence of a catalyst.

KClO3 KCl + O2 (unbalanced)
What volume of O2 gas will be generated at T = 35°C and P = 764.0 torr from 1.51 g of KClO3?
I got .414 L and used 22.4 L for moles of O2
What's going on??

22.4 L/mol is the volume at STP and you don't have STP conditions.

Balance the equation, find n O2, then use PV = nRT and substitute the conditions of P and T listed (remember T must be in kelvin) and solve for V in L.

I would convert torr to atm (torr/760) and use 0.08206 for R.

Ok thanks for your help!

You're welcome.

To calculate the volume of oxygen gas that will be generated, we need to follow a few steps:

Step 1: Convert the mass of KClO3 to moles.
The molar mass of KClO3 is 122.55 g/mol. To convert grams to moles, use the formula: moles = mass / molar mass.
In this case, moles = 1.51 g / 122.55 g/mol ≈ 0.01233 mol

Step 2: Determine the stoichiometry of the balanced equation.
The balanced equation shows that for every one mole of KClO3, one mole of O2 gas is produced.

Step 3: Use the ideal gas law to calculate the volume of O2 gas.
The ideal gas law is expressed as PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the gas constant, and T is the temperature in Kelvin. In this case, we are given the pressure (764.0 torr) and temperature (35°C) and need to solve for the volume.

Convert the temperature from Celsius to Kelvin by adding 273.15:
T = 35°C + 273.15 = 308.15 K

Plug the values into the ideal gas law equation:
PV = nRT
V = (nRT) / P

V = (0.01233 mol * 0.0821 L*atm/mol*K * 308.15 K) / 764.0 torr

Now, we need to convert torr to atm, since the gas constant is in units of atm:
1.00 atm = 760 torr, so 764.0 torr = 764.0/760 ≈ 1.005

V = (0.01233 mol * 0.0821 L*atm/mol*K * 308.15 K) / 1.005 atm ≈ 0.379 L

So, the volume of O2 gas generated at T = 35°C and P = 764.0 torr from 1.51 g of KClO3 is approximately 0.379 L.

Your calculation of 0.414 L might have been due to using the value of 22.4 L as a conversion factor for moles of O2 gas, which is only true under standard temperature and pressure (STP) conditions (0°C or 273.15 K and 1.00 atm). In this case, we are given different temperature and pressure values, so the ideal gas law equation needs to be used.