18. Benzene (CoHo) undergoes combustion upon reaction with oxygen i) Balance the

Question

18. Benzene (CoHo) undergoes combustion upon reaction with oxygen i) Balance the equation ii) Fill in the following table. In each row, you are given an amount of both reactants. Determine which of them is the limiting reactant, and then calculate the amounts of the products in the units indicated. Show all your calculations in the space below, or attach an additional page. Amount of Amount of O2 Lim. Amount of CO2 Amount of CsH6 Reactant produced water produced | |- | C.re A 11 moles 45 moles molesmoles | -4-moles -V-moles 7.0 moles 54.0 moles cl 14.5 grams | 19.3 grams 10% grams ms Calculations: A) C)

Explanation / Answer

2 C6H6 + 15 O2 ----> 12 CO2 + 6 H2O

2 mol of Benzene reacts with 15 mol of O2 to produce 12 mol of CO2 and 6 mol of H2O

A) 11 mol of C6H6   + 45 mol of O2

for complete reaction of 11 mol C6H6 = 11 mol C6H6 * ( 15 mol O2 / 2 mol C6H6) => 82.5 mol of O2

but only 45 mol of O2 is available

so 45 mol O2 * ( 2 mol C6H6 / 15 mol O2) => 6 mol of C6H6 can react ( reamainig 11 - 6 = 5 mol of C6H6 wiil be unreacted)

Hence limitting reactant is O2

amount of CO2 = 6 mol C6H6 * ( 12 mol CO2 / 2 mol C6H6) => 36 mol of CO2

amount of H2O = 6 mol C6H6 * ( 6 mol H2O / 2 mol C6H6) => 18 mol of H2O

C) 14.5 g of C6H6 = 14.5 g / 78.11 g/mol => 0.1856 mol of C6H6   ( molar mass of C6H6 = 78.11 g/mol)

19.3 of O2 = 19.3 g / 32 g/mol => 0.6031 mol of O2 ( molar mass of O2 = 32 g/mol)

for complete reaction of 0.1856 mol C6H6 =  0.1856 mol C6H6 * ( 15 mol O2 / 2 mol C6H6) => 1.392 mol of O2

but only 0.6031 mol of O2 is available

so 0.6031 mol O2 * ( 2 mol C6H6 / 15 mol O2) => 0.0804 mol of C6H6 can react ( reamainig 0.1856 - 0.0804 =0.1052 mol of C6H6 wiil be unreacted)

Hence limitting reactant is O2

amount of CO2 = 0.0804 mol C6H6 * ( 12 mol CO2 / 2 mol C6H6) =>0.4824 mol of CO2

molar mass of CO2 = 44 g/mol

0.4824 mol * 44 g/mol => 21.2 g of CO2

amount of H2O = 0.0804 mol C6H6 * ( 6 mol H2O / 2 mol C6H6) => 0.2412 mol of H2O

Molar mass of H2O = 18 g/mol

0.2412 mol * 18 g/mol => 4.34 g of H2O

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