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:

$\begin{gathered} n=\frac{PV}{RT} \\ n=\frac{0.9801atm\times0.013L}{0.08206\frac{atm.L}{mol.K}\times295.15K} \\ n=\frac{0.9801\times0.013}{0.08206\times295.15}mol=5.3\times10^{-4}molO_2 \end{gathered}$

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

$\begin{gathered} gO_2=GivenmolO_2\times\frac{MolarMass,gO_2}{1molO_2} \\ gO_2=5.3\times10^{-4}molO_2\times\frac{31.999gO_2}{1molO_2}=0.0168gO_2=1.7\times10^{-2}gO_2 \end{gathered}$

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