1.) When 25.0 g of ethanol and 55.0 g of oxygen combine in the combustion reacti

Question

1.) When 25.0 g of ethanol and 55.0 g of oxygen combine in the combustion reaction: 2C3H7OH (l) + 9 O2 (g) --> 6CO2 (g) + 8H2O (g)

a) whats the limiting reactant and what is the theoretical yield of Carbon Dioxide

b)What is the excess reactant and how many grams of it remains at the end of the reaction?

2.) The most important commerical process for converting N2 from the air into nitrogen-containing compunds in based on the reaction of N2 and H2 to form ammonia: (NH3): N2 (g) + 3 H2 (g) --->2 NH3 (g)

a.) how many moles of NH3 can be formed from 15.0 mol of N2 and 15.0 mol of H2?

Explanation / Answer

Number of moles of ethanol in 55 g = 25/59 = 0.42373 mol

Number of moles of oxygen in 55/32 =1.71875 mol

2 moles ethanol requires 9 mole oxygen

1 mole ethanol requires 4.5 mol oxygen

thus 0.42373 require 0.42373*4.5 mol oxygen = 1.90679 mol

But we have only 1.71875 mol oxygen. Thus OXYGEN is the limiting reactant.

Now

2 mol oxygen forms 6 mole CO2

thus 1,71875 mole O2 forms 3*1.71875 mol = 5.15625 ml CO2

Now 1.7185 mol O2 will use 1.71875/4.5 mol ethanol =0.38194

thus amount left = 0.42373-0.38194 =0.4179 mol=24.6561 gram

here H2 is the limiting reactant(similarly)

2 mole H2 give 3 mole ammonia

thus 15 mol will give 1.5*15 = 2.25 mole ammonia

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