## General, Organic, and Biological Chemistry: Structures of Life (5th Edition)

a. $$C_2H_6O + 3O_2 \longrightarrow 2CO_2 + 3H_2O$$ b. $24 \space moles \space O_2$ c. 4.8 g of $O_2$ d. $239 \space g \space CO_2$ and $147 \space g \space H_2O$
a. Identify each reactant and product, and write the unbalanced equation: Reactants: Ethanol $(C_2H_6O)$ and Oxygen $(O_2)$ Products: Carbon dioxide $(CO_2)$ and Water $(H_2O)$ $$C_2H_6O + O_2 \longrightarrow CO_2 + H_2O$$ - Balance the number of carbons: $$C_2H_6O + O_2 \longrightarrow 2CO_2 + H_2O$$ - Balance the number of hydrogens: $$C_2H_6O + O_2 \longrightarrow 2CO_2 + 3H_2O$$ - Balance the number of oxygens: $$C_2H_6O + 3O_2 \longrightarrow 2CO_2 + 3H_2O$$ b. $$8.0 \space moles \space C_2H_6O \times \frac{ 3 \space moles \ O_2 }{ 1 \space mole \space C_2H_6O } = 24 \space moles \space O_2$$ c. $CO_2$ : ( 12.01 $\times$ 1 )+ ( 16.00 $\times$ 2 )= 44.01 g/mol $$\frac{1 \space mole \space CO_2 }{ 44.01 \space g \space CO_2 } \space and \space \frac{ 44.01 \space g \space CO_2 }{1 \space mole \space CO_2 }$$ $O_2$ : ( 16.00 $\times$ 2 )= 32.00 g/mol $$\frac{1 \space mole \space O_2 }{ 32.00 \space g \space O_2 } \space and \space \frac{ 32.00 \space g \space O_2 }{1 \space mole \space O_2 }$$ $$4.4 \space g \space CO_2 \times \frac{1 \space mole \space CO_2 }{ 44.01 \space g \space CO_2 } \times \frac{ 3 \space moles \space O_2 }{ 2 \space moles \space CO_2 } \times \frac{ 32.00 \space g \space O_2 }{1 \space mole \space O_2 } = 4.8 \space g \space O_2$$ d. $C_2H_6O$ : ( 1.008 $\times$ 6 )+ ( 12.01 $\times$ 2 )+ ( 16.00 $\times$ 1 )= 46.07 g/mol $$\frac{1 \space mole \space C_2H_6O }{ 46.07 \space g \space C_2H_6O } \space and \space \frac{ 46.07 \space g \space C_2H_6O }{1 \space mole \space C_2H_6O }$$ $CO_2$ : 44.01 g/mol $$\frac{1 \space mole \space CO_2 }{ 44.01 \space g \space CO_2 } \space and \space \frac{ 44.01 \space g \space CO_2 }{1 \space mole \space CO_2 }$$ $$125 \space g \space C_2H_6O \times \frac{1 \space mole \space C_2H_6O }{ 46.07 \space g \space C_2H_6O } \times \frac{ 2 \space moles \space CO_2 }{ 1 \space mole \space C_2H_6O } \times \frac{ 44.01 \space g \space CO_2 }{1 \space mole \space CO_2 } = 239 \space g \space CO_2$$ $H_2O$ : ( 1.008 $\times$ 2 )+ ( 16.00 $\times$ 1 )= 18.02 g/mol $$\frac{1 \space mole \space H_2O }{ 18.02 \space g \space H_2O } \space and \space \frac{ 18.02 \space g \space H_2O }{1 \space mole \space H_2O }$$ $$125 \space g \space C_2H_6O \times \frac{1 \space mole \space C_2H_6O }{ 46.07 \space g \space C_2H_6O } \times \frac{ 3 \space moles \space H_2O }{ 1 \space mole \space C_2H_6O } \times \frac{ 18.02 \space g \space H_2O }{1 \space mole \space H_2O } = 147 \space g \space H_2O$$