[Lawsen Lew]

CO2 44.0098 g/mol
SO2 64.0648 g/mol
We could use substitution algebra to solve this problem. I guess these gases will not react together or leak out. I checked another reply and learned this algebraic technique. Sorry for any errors. I am not an expert chemist or mathematician. Who says forums are reliable or absolute, that is why you need to be careful of me. I am not always correct about the world. I try my best to answer your question out of curiosity. I recently graduated with nothing to do, but surf the web. It is 2:55 AM and I am not obligated to solve your chemistry problems. I am not in school anymore, nothing matters and it will not be long before my mom will claim all my things back, repatriation of my separate home while attending school.
Let x= mole CO2
Let y=mole SO2

44.0098X+64.0648Y=2.952 g mixture

X + Y = 0.05300 mol mixture
Solve for Y to substitute it back into the equation to solve for X to get the mole of CO2
-x+X + Y=0.05300-x
Y=0.05300-X

44.0098X+64.0648(0.05300-X)=2.952 g mixture
44.0098X+3.39543-64.0648X=2.952 g mixture
3.39543-20.055X=2.952 g mixture
-20.055X=-0.44343
x=-0.44343/-20.055
x=0.022111 mol CO2 (44.0098 g CO2/ 1 mol CO2) = 0.973101 g CO2

Y=0.05300-X
Y=0.05300 mol mixture-0.022111 mol CO2
Y=0.030889 mol SO2 (64.0648g/mol SO2) = 1.9789 g SO2

Checking the answer by totaling the masses of gases:
1.9789 g SO2 + 0.973101 g CO2 = 2.952 g gases mixture of CO2 and SO2
Conclusion: There is 0.9731 g CO2 gas in this mixture in your place to use.

This is a tricky question, I am not sure if I am doing it correctly. I am BS in earth science geology at San Jose State U. Typical state CSU schools out of the other schools on the planet, but that is how much my mom can afford, no loans and no debts. This one is located in the Santa Clara valley region in down town San Jose, not far from several major universities.- turtle