This is a limiting ragent problem. I know that because amounts are given for BOTH reactants.
When dealing with gases and the T and P remain the same, one can take a short cut ans use volume directly as if volume = mols. It saves a little time.
a. Calculate volume butane formed with 17.2 L CO and all the H2 needed. That's 17.2 L CO x (1 mol C4H10/4 mols CO) = 17.2/4 = approx 4.30 L.
b. Calculate volume butane formed with 42.1L H2 and all the CO needed. That's
42.1 L H2 x (1 mol C4H10/9 mols H2) = approx 4.7.
c. The values for L C4H10 don't agree which means one is wrong; the correct value in limiting reagent problems is ALWAYS the smaller value and the reagent responsible for that is the limiting reagent.
d. That means CO is the limiting reagent and H2 is the excess reagent.
e. How much H2 will be used up with the CO? That's 17.2 L CO x (9 mol H2/4 mol CO) = about 38.7 L H2 used.
f. We started with 42.1 LH2; we used 38.7 L H2, the difference is the amount H2 not used.
Carbon monoxide, CO and hydrogen, H2, react accordingly:
4CO(g)+9H2(g)-->C4H10(g)+4H2O
What volume of the EXCESS REACTANT remains if 17.2L CO and 42.1L H2 are allowed to react?
Assume both gases are measured at 711 degree celsius and 1.33atm.
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1 answer