Mmmmmmmmmmmmmmmmmmmjjjjjjjj
<span>2.10 grams.
The balanced equation for the reaction is
CO + 2H2 ==> CH3OH
The key thing to take from this equation is that it takes 2 hydrogen molecules per carbon monoxide molecule for this reaction. And since we've been given an equal number of molecules for each reactant, the limiting reactant will be hydrogen.
We can effectively claim that we have 5.86/2 = 2.93 l of hydrogen and an excess of CO to consume all of the hydrogen. So the number of moles of hydrogen gas we have is:
2.93 l / 22.4 l/mol = 0.130803571 mol
And since it takes 2 moles of hydrogen gas to make 1 mole of methanol, divide by 2, getting.
0.130803571 mol / 2 = 0.065401786 mol
Now we just need to multiply the number of moles of methanol by its molar mass. First lookup the atomic weights involved.
Atomic weight carbon = 12.0107 g/mol
Atomic weight hydrogen = 1.00794 g/mol
Atomic weight oxygen = 15.999 g/mol
Molar mass CH3OH = 12.0107 + 4 * 1.00794 + 15.999 = 32.04146 g/mol
So the mass produced is
32.04146 g/mol * 0.065401786 mol = 2.095568701 g
And of course, properly round the answer to 3 significant digits, giving 2.10 grams.</span>
From the ideal gas law
pv=nRT , n is therefore PV/RT
R is the
R is gas constant =62.364 torr/mol/k
P=500torr
V=4.00l
T=500+273=773k
n={(500 torr x 4.00l)/(62.364 x773k)}=0.041moles
the number of molecules=moles x avorgadro costant that is 6.022x10^23)
6.022 x 10^23) x0.041=2.469 x10^22molecules
Carbon (after hydrogen, helium and oxygen).
Answer:
161.57 K
Explanation:
V1/T1 = v2/T2
(5.24 L)/(411 K) = (2.06 L)/T2
T2 = 161.57 K
