Question

What mass of hydrogen is present in a mixture of `H_2`, `CO`, and `O_2` at `765*mm*Hg` at `23.8` `""^@C`, that contains `8.0*g` `CO`, and `12.0*g` `O_2`?

Answer 1

Well, first let us determine the TOTAL molar quantity of each gas by means of the Ideal Gas Equation....and a get approx. `600*mg` mass with respect to dihydrogen......

Explanation:

`n_"Total"=(PV)/(RT)=((765*mm*Hg)/(760*mm*Hg*atm^-1)xx23.8*L)/(0.0821*(L*atm)/(K*mol)xx297.0*K)=0.982*mol`

Now `0.982*mol=n_"CO"+n_"dioxygen"+n_"dihydrogen"`

And `n_"CO"=(8.00*g)/(28.01*g*mol^-1)=0.286*mol`.

`n_(O_2)=(12.00*g)/(32.01*g*mol^-1)=0.375*mol`.

`n_"dihydrogen"=n_"Total"-n_"dioxygen"-n_"carbon monoxide"`.

`=(0.982-0.375-0.286}*mol=0.321*mol`..

And thus a mass of `0.321*molxx20.015*g*mol^-1=??*g` with respect to `"dihydrogen."`

Answer 2

The mass of hydrogen present is 0.650 g.

Explanation:

Step 1. Calculate the total moles of gas

We can use the Ideal Gas Law.

`color(blue)(bar(ul(|color(white)(a/a)bb((1))color(white)(m)pV = nRTcolor(white)(a/a)|)))" "`

So

`bb((2))color(white)(m)n = (pV)/(RT)`

#p_text(tot) = 765 color(red)(cancel(color(black)("mmHg"))) × "1 atm"/(760 color(red)(cancel(color(black)("mmHg")))) = "1.007 atm"#

`T = "(23.8 + 273.15) °C = 296.95 °C"`

`n_text(tot) = (1.007 color(red)(cancel(color(black)("atm"))) × 23.8 color(red)(cancel(color(black)("L"))))/("0.082 06" color(red)(cancel(color(black)("L·atm·K"^"-1")))"mol"^"-1" × 296.95 color(red)(cancel(color(black)("K")))) = "0.9831 mol"`

Step 2. Calculate the moles of `"H"_2`

`n_text(CO) = 8.00 color(red)(cancel(color(black)("g CO"))) × "1 mol CO"/(28.01 color(red)(cancel(color(black)("g CO")))) = "0.2856 mol CO"`

`n_text(O₂) = 12.00 color(red)(cancel(color(black)("g O"_2))) × "1 mol O"_2/(32.00 color(red)(cancel(color(black)("g O"_2)))) = "0.3750 mol O"_2`

`n_text(tot) = n_text(CO) + n_text(O₂) + n_text(H₂)`

`n_text(H₂) = n_text(tot) - n_text(CO) - n_text(O₂)`

∴ `n_text(H₂) = "(0.9831 - 0.2856 - 0.3750) mol = 0.3225 mol"`

Step 3. Calculate the mass of `"H"_2`

`"Mass of H"_2 = 0.3225 color(red)(cancel(color(black)("mol H"_2))) × "2.016 g H"_2/(1 color(red)(cancel(color(black)("mol H"_2)))) = "0.650 g H"_2`