Question

How many moles of $C{O_2}$ form when 58.0 g of butane, ${C_4}{H_{10}}$ burns in oxygen?

Answer 1

Hint: Butane reacts with oxygen to form carbon dioxide and water. From the reaction the mole ratio of carbon dioxide and butane is obtained. The number of moles of butane is calculated by dividing the mass of the butane with the molecular weight of the butane.

Complete step by step answer:
It is given that the mass of butane is 58.0 g.
In the question, butane burns in presence of oxygen to form carbon dioxide.
The reaction is shown below.
${C_4}{H_{10}} + {O_2} \to C{O_2} + {H_2}O$
In this reaction, butane reacts with oxygen to form carbon dioxide and water.
The balanced chemical equation is shown below.
$2{C_4}{H_{10}} + 13{O_2} \to 8C{O_2} + 10{H_2}O$
In this reaction, two mole of butane reacts with thirteen mole of oxygen to form eight mole of carbon dioxide and ten mole of water.
The molecular weight is calculated by adding the atomic weight of the atom multiplied with the number of atoms present.
The atomic weight of carbon is 12.011 g/mol.
The atomic weight of hydrogen is 1.008 g/mol.
The molecular weight of butane is calculated as shown below.
$ \Rightarrow molecular\;weight = (4 \times 12.011 + 10 \times 1.008)$
$ \Rightarrow molecular\;weight = 58.124g/mol$
From the given reaction we know that 2 mole of butane gives 8 mole of carbon dioxide. So, the mole ratio is given as,
$ \Rightarrow \dfrac{{8\;mol\;of\;C{O_2}}}{{2\;mol\;of\;{C_4}{H_{10}}}}$
The formula to calculate the number of moles is shown below.
$n = \dfrac{m}{M}$
Where,
n is the number of moles
m is the mass
M is the molar mass
To calculate the number of moles, substitute the value in the above equation.
$ \Rightarrow n = \dfrac{{58.0g}}{{58.124g/mol}}$
$ \Rightarrow n = 0.99mol \approx 1mol$
To calculate the moles of carbon dioxide multiply the number of moles of butane with the mole ratio.
$ \Rightarrow 1mol \times \dfrac{{8mol\;C{O_2}}}{{2mol\;{C_4}{H_{10}}}}$
$ \Rightarrow 4mol$
Therefore, 4 moles of $C{O_2}$ form when 58.0 g of butane, ${C_4}{H_{10}}$ burns in oxygen.

Note:
Make sure to balance the reaction because if not balanced we will not get the actual mole ratio which helps to calculate the moles of carbon dioxide.