Question

Which contains more molecules? 10.0 g of CO2 or 5.60L of oxygen gas at STP

Answer 1

Oxygen.

Step-by-step explanation:

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In this case, since the ideal gas equation allows us to compute the moles of oxygen in 5.60 L at STP (1 atm and 273.15 K) as shown below:

`PV=nRT\\\\n=(PV)/(RT)=(1atm*5.60L)/(0.08206(atm*L)/(mol*K)*273.15K)\\\\n_(O_2)= 0.25mol`

Next, given the molar mass of carbon dioxide (44.01 g/mol) we compute the moles in 10.0g of this gas via:

`n_(CO_2)=10.0gCO_2*(1molCO_2)/(44.01gCO_2) =0.23molCO_2`

Thus, since oxygen has the greatest number of moles, we immediately infer it also has the greatest number of molecules based on the Avogadro's number.

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