Answer
46.64 g of water.
Work Step by Step
This is a combustion reaction, thus, $C_2H_5OH$ will react with $O_2$ to produce carbon dioxide and water:
$$C_2H_5OH + O_2 \longrightarrow CO_2 + H_2O$$
- Balance the reaction:
$$C_2H_5OH + 3O_2 \longrightarrow 2CO_2 + 3H_2O$$
- 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:
$$ 100.0 \space g \times \frac{1 \space mole}{ 46.07 \space g} = 2.171 \space moles$$
- 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:
$$ 82.82 \space g \times \frac{1 \space mole}{ 32.00 \space g} = 2.588 \space moles$$
Find the amount of product if each reactant is completely consumed.
$$ 2.171 \space moles \space C_2H_5OH \times \frac{ 3 \space moles \ H_2O }{ 1 \space mole \space C_2H_5OH } = 6.513 \space moles \space H_2O $$
$$ 2.588 \space moles \space O_2 \times \frac{ 3 \space moles \ H_2O }{ 3 \space moles \space O_2 } = 2.588 \space moles \space H_2O $$
Since the reaction of $ O_2 $ 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:
$$ 2.588 \space mole \times \frac{ 18.02 \space g}{1 \space mole} = 46.64 \space g$$
