Question

This is my data- Water temp.: 22 C, Pressure: 0.9801 atm, Final volume: 13 mL . The question is: Using the Ideal Gas Law (PV = nRT), calculate the grams of O2 produced in the reaction. (Hint: solve for n, and then convert moles to grams. Don’t forget to convert your temperature from Celsius to Kelvin.) Show your work. This was the answer I got: 0.01676 grams

Answer 1

We are told that we must assume that the gas behaves as an ideal gas in order to apply the following equation:

`PV=nRT`

Where,

P is the pressure of the gas, 0.9801atm

V is the volume of the gas, 13mL=0.013L

R is a constant, 0.08206 atm-L/mol-K

T is the temperature of the gas, 22°C=295.15K

Now, we will clear the moles of the gas, n and we replace the known data:

`\begin{gathered} n=(PV)/(RT) \\ n=(0.9801atm*0.013L)/(0.08206(atm.L)/(mol.K)*295.15K) \\ n=(0.9801*0.013)/(0.08206*295.15)mol=5.3*10^(-4)molO_2 \end{gathered}`

now, to have the mass of the gas, we will use the molar mass. In this case, the molar mass of O2=31.999g/mol

`\begin{gathered} gO_2=GivenmolO_2*(MolarMass,gO_2)/(1molO_2) \\ gO_2=5.3*10^(-4)molO_2*(31.999gO_2)/(1molO_2)=0.0168gO_2=1.7*10^(-2)gO_2 \end{gathered}`

The grams of O2 produced in the reaction are 1.7x10^-2 g of O2