Answer
(a) The mixture of reactants will have 8 CO molecules and 6 $O_2$ molecules.
(b) The product mixture has 2 $O_2$ molecules and 8 $CO_2$ molecules.
(c) 211.6 g of $CO_2$
Work Step by Step
(a) The reactants mixture is made of 8 CO molecules and 6 $O_2$ molecules, as said in the exercise.
(b) 8 CO molecules can produce 8 $CO_2$ molecules.
6 $O_2$ molecules can produce 12 $CO_2$ molecules.
Thus, $CO$ is the limiting reactant, and 8 $CO_2$ molecules will be produced.
To produce 8 $CO_2$ molecules, we need 4 $O_2$. So, 2 oxygen molecules will remain in the products.
(c)
- Calculate or find the molar mass for $ CO $:
$ CO $ : ( 12.01 $\times$ 1 )+ ( 16.00 $\times$ 1 )= 28.01 g/mol
- Using the molar mass as a conversion factor, find the amount in moles:
$$ 134.67 \space g \times \frac{1 \space mole}{ 28.01 \space g} = 4.808 \space moles$$
- 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:
$$ 77.25 \space g \times \frac{1 \space mole}{ 32.00 \space g} = 2.414 \space moles$$
Find the amount of product if each reactant is completely consumed.
$$ 4.808 \space moles \space CO \times \frac{ 2 \space moles \ CO_2 }{ 2 \space moles \space CO } = 4.808 \space moles \space CO_2 $$
$$ 2.414 \space moles \space O_2 \times \frac{ 2 \space moles \ CO_2 }{ 1 \space mole \space O_2 } = 4.828 \space moles \space CO_2 $$
Since the reaction of $ CO $ produces less $ CO_2 $ for these quantities, it is the limiting reactant.
$ CO_2 $ : ( 12.01 $\times$ 1 )+ ( 16.00 $\times$ 2 )= 44.01 g/mol
- Using the molar mass as a conversion factor, find the mass in g:
$$ 4.808 \space mole \times \frac{ 44.01 \space g}{1 \space mole} = 211.6 \space g$$
