Posted by Mark on Monday, July 16, 2012 at 10:19pm.
You must first determine which gas is in excess; i.e., which is the limiting reagent.
2H2 + O2 ==> 2H2O
mol H = g/molar mass = 5/2 = 2.5
mol O = 10/32 = 0.31
Using all of the H2 gas and all of the oxygen needed, how much H2O will be formed? That is
2.5 mol H2 x (2 mol H2O/2 mol H2) = 2.5 x 2/2 = 2.5 mol H2O form3ed.
Using all of the O2 gas and all of the hydrogen needed, how much H2O will be formed? That is
0.31 x (2 mol H2O/1 mol O2) = 0.31 mol H2O.
Obviously both of these answers can't be correct; therefore, in limiting reagent problems the SMALLER amount is the correct one and the reagent producing that value is the limiting reagent. Thus, O2 is the limiting reagent and hydrogen is in excess.
Next you want to determine how much H2 is used. That is
0.31 mol O2 x (2 mol H2/1 mol O2) = 0.31 x 2/1 = 0.62
We had 2.5 initially, we've used 0.62; therefore, we have 2.5-0.62 = about 1.9 remain unreacted. Use PV = nRT and solve for P (in atmospheres) using 298 for T in kelvin. I get an answer of about 3 atm. Check my work.
How many moles of hydrogen atoms are present in 2.41x10^24 molecules of CH4?
See your other question on CH3OH. The precess is the same.
DrBob222 -
You're the greatest!!!! Thank you!