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Sagot :
We are told that we must assume that the gas behaves as an ideal gas in order to apply the following equation:
[tex]PV=nRT[/tex]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:
[tex]\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}[/tex]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
[tex]\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}[/tex]The grams of O2 produced in the reaction are 1.7x10^-2 g of O2
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