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#?

2 Answers
Nov 23, 2017

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."#

Nov 23, 2017

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#