8.26 g heptane and 8.43 g oxygen react according to the followingunbalanced equa

Question

8.26 g heptane and 8.43 g oxygen react according to the followingunbalanced equation:

C7H16 + 11 O2 8 H2O + 7CO2


Please help with the 3 questions below:

Q1: Calculate the theoretical yield (ingrams) of CO2?

Q2: Calculate the mass in grams of the excessreagent remaining after the complete reaction ofC7H16 with of O2?

Q3: If 4.28 g ofCO2 is actually produced, determine the percent yieldfor the reaction?



Thanks a lot !

Explanation / Answer

First, you need to determine which reactant is limitedreactant. By: Mole of heptane: 8.26g/100.198g/mol=0.0824mol        Mole of oxygen:8.43g/32g/mol=0.263(mol) From reaction, mole ratio between heptane and oxygen is 1:11               C7H16 + 11 O2 8 H2O + 7CO2 I    0.0824    0.263    x       -11x E         0.0824-x     0.263-11x If heptane is limited reactant=> 0.0824-x = 0=> x =0.0824     Oxygen is -------------   =>0.263*11x = 0=>x = 0.0239mol          Since0.0239 Oxygen is limitedreactant a) use O2 to find CO2         In reaction, ratiobetween O2 and CO2 is 11:7        => mole of CO2: 0.263*7/11=0.167mol          mass of CO2:0.167*44.01=7.366g b) Mole of heptane reacts with O2: 0.263/11 = 0.0239mol    Mole of excess heptane: 0.0824-0.0239 =0.0585(mol)    mass of excess heptane: 0.0585*100.198g/mol =5.86g c)%yield : 4.28g/7.366g *100 =58.1%

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