Chemistry: Molecular Approach (4th Edition)

Limiting reactant: Ethanol $(C_2H_5OH)$ Theoretical yield: $4.28$ g Percent yield: 87.0 %
1. Find the masses of ethanol and water: $$4.62 \space mL \space C_2H_5OH \times \frac{0.789 \space g}{1 \space mL} = 3.64\underline{52} \space g \space C_2H_5OH$$ $$3.72 \space mL \space H_2O \times \frac{1.00 \space g}{1 \space mL} =3.72 \space g \space H_2O$$ 2. This is a combustion reaction; ethanol will react with oxygen to produce water and carbon dioxide. $$C_2H_5OH + O_2 \longrightarrow CO_2 + H_2O$$ - Balance the amount of carbon atoms, then the amount of hydrogen, and the amount of oxygen atoms. $$C_2H_5OH + \frac 72 O_2 \longrightarrow 2CO_2 + 3H_2O$$ - Multiply all coefficients by 2 to remove the fraction: $$2C_2H_5OH + 7O_2 \longrightarrow 4CO_2 + 6H_2O$$ 3. Find the limiting reactant, and calculate the theoretical and percent yield. - Calculate or find the molar mass for $C_2H_5OH$: $C_2H_5OH$ : ( 1.008 $\times$ 6 )+ ( 12.01 $\times$ 2 )+ ( 16.00 $\times$ 1 )= 46.07 g/mol - Using the molar mass as a conversion factor, find the amount in moles: $$3.6452 \space g \times \frac{1 \space mole}{ 46.07 \space g} = 0.07912 \space mole$$ - Calculate or find the molar mass for $O_2$: $O_2$ : ( 16.00 $\times$ 2 )= 32.00 g/mol - Using the molar mass as a conversion factor, find the amount in moles: $$15.55 \space g \times \frac{1 \space mole}{ 32.00 \space g} = 0.4859 \space mole$$ Find the amount of product if each reactant is completely consumed. $$0.07912 \space mole \space C_2H_5OH \times \frac{ 6 \space moles \ H_2O }{ 2 \space moles \space C_2H_5OH } = 0.2374 \space mole \space H_2O$$ $$0.4859 \space mole \space O_2 \times \frac{ 6 \space moles \ H_2O }{ 7 \space moles \space O_2 } = 0.4165 \space mole \space H_2O$$ Since the reaction of $C_2H_5OH$ produces less $H_2O$ for these quantities, it is the limiting reactant. - Calculate or find the molar mass for $H_2O$: $H_2O$ : ( 1.008 $\times$ 2 )+ ( 16.00 $\times$ 1 )= 18.02 g/mol - Using the molar mass as a conversion factor, find the mass in g: $$0.2374 \space mole \times \frac{ 18.02 \space g}{1 \space mole} = 4.27\underline{8} \space g = 4.28 \space g$$ $$Percent \space yield = \frac{ 3.72 }{ 4.278 } \times 100\% = 87.0 \%$$