The chemical formula for benzene is "C"_6"H"_6"C6H6. When benzene undergoes combustion, carbon dioxide and water are formed.
Balanced Equation
"2C"_6"H"_6 + "15O"_2"2C6H6+15O2rarr→"12CO"_2 + "6H"_2"O"12CO2+6H2O
The process for answering this question will be:
color(blue)"given mass benzene"given mass benzenerarr→color(blue)"mol benzene"mol benzenerarr→color(blue)"mol oxygen"mol oxygenrarr→color(blue)"mass oxygen"mass oxygen
In order to do this, we will need the appropriate mole ratio from the balanced equation, and the molar masses of benzene and oxygen.
We need the mole ratio between benzene and oxygen from the balanced equation.
Mole Ratio Benzene: Oxygen
("2 mol C"_6"H"_6)/("15 mol O"_2")2 mol C6H615 mol O2 and ("15 mol O"_2)/("2 mol C"_6"H"_6")15 mol O22 mol C6H6
Molar Masses of Benzene and Oxygen
Multiply the subscript of each element by its molar mass, the atomic weight on the periodic table in g/mol.
"C"_6"H"_6":C6H6:(6xx12.011"g/mol C")+(6xx1.008"g/mol H")="78.114 g/mol C"_6"H"_6"(6×12.011g/mol C)+(6×1.008g/mol H)=78.114 g/mol C6H6
"O"_2":O2:(2xx15.999"g/mol O")="31.998 g/mol O"_2"(2×15.999g/mol O)=31.998 g/mol O2
Determine Moles Benzene
Multiply the color(blue)"given mass"given mass of benzene by the reciprocal of its molar mass.
150.0color(red)cancel(color(black)("g C"_6"H"_6))xx(1"mol C"_6"H"_6)/(78.114color(red)cancel(color(black)("g C"_6"H"_6)))="1.9203 mol C"_6"H"_6"
Determine Moles Oxygen Using the Mole Ratio
1.9203 color(red)cancel(color(black)("mol C"_6"H"_6))xx("15 mol O"_2)/(2 color(red)cancel(color(black)("mol C"_6"H"_6)))="14.402 mol O"_2"
Determine Mass Oxygen Using Molar Mass of Oxygen
Multiply mol oxygen by its molar mass.
14.402color(red)cancel(color(black)("mol O"_2))xx(31.998"g O"_2)/(1color(red)cancel(color(black)("mol O"_2)))="460.8 g O"_2" rounded to four significant figures
