Respuesta :
Answer: 1.95 g of oxygen is present.
Explanation:
According to Dalton's law, the total pressure is the sum of individual pressures.
[tex]p_{total}=p_{methane}+p_{oxygen}[/tex]
Given : [tex]p_{total}[/tex] =total pressure of gases = 727 mm Hg
[tex]p_{methane}[/tex] = partial pressure of methane = 571 mm Hg
[tex]p_{oxygen}[/tex] = partial pressure of oxygen = ?
[tex]727=571+p_{oxygen}[/tex]
[tex]p_{oxygen}=156mmHg[/tex]
Also [tex]p_{oxygen}=x_{oxygen}\times p_{total}[/tex]
Given : 3.62 g of methane is present
moles of methane = [tex]\frac{\text{Given mass}}{\text {Molar mass}}=\frac{3.62g}{16g/mol}=0.226moles[/tex]
[tex]x_{oxygen}[/tex] = mole fraction of oxygen
=[tex]\frac{\text {moles of oxygen}}{\text {total moles}}=\frac{y}{y+0.226}[/tex]
[tex]156=\frac{y}{y+0.226}\times 727[/tex]
[tex]y=0.061[/tex]
mass of oxygen = [tex]moles\times {\text {Molar mass}}=0.061\times 32=1.95g[/tex]
Thus 1.95 g of oxygen is present.
