Question #12a91

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Answer 1 · 2017-05-16T01:49:52

#2.07gC_3H_8#

First, let's find the moles of #O_2# using the ideal-gas equation:

#molO_2 = ((1.04atm)(5.53L))/((0.08206 (L-atm)/(mol-K))(298K)) = 0.235 molO_2#

Now, we'll use the stoichiometrically equivalent values from the equation to find the relative number of moles of #C_3H_8#:

#0.235molO_2((1 molC_3H_8)/(5 mol O_2)) = 0.0470molC_3H_8#

Finally, we'll use the molar mass of propane to find the mass in grams of propane needed:

#0.0470molC_3H_8((44.11gC_3H_8)/(1 mol C_3H_8)) = 2.07gC_3H_8#