Complete Question
What volume of butane can be produced from the reaction of 13.45 g of carbon and 17.65 L of hydrogen gas at STP ?
Answer:
The volume of butane produced is [tex]V_b = 3.54 \ L[/tex]
Explanation:
From the question we are told that
The mass of carbon is [tex]m_c = 13.45 \ g[/tex]
The volume of hydrogen is [tex]V_h = 17.65\ L[/tex]
The number of moles of carbon is mathematically evaluated as
[tex]n_c = \frac{m_c}{M_c}[/tex]
Where [tex]M_c[/tex] is the molar mass of carbon which is a constant with value [tex]M_c = 12\ g/mol[/tex]
So
[tex]n_c = \frac{13.45}{12}[/tex]
[tex]n_c = 1.12[/tex]
Now since the volume of the hydrogen is measured at standard temperature and pressure
Hence the number of moles of hydrogen is
[tex]n_h = \frac{V_h }{22.4}[/tex]
Where 22.4 is a constant
[tex]n_h = \frac{17.65}{22.4}[/tex]
[tex]n_h = 0.79[/tex]
Comparing [tex]n_h \ and\ n_c[/tex] we see that hydrogen is the limiting reactant
because [tex]n_h[/tex] is greater than [tex]n_c[/tex]
The chemical equation for this reaction is
[tex]4C_{(s)} + 5H_2-{(g)} ----> C_5 H_10_{(g)}[/tex]
looking at chemical equation we see that
5 moles of hydrogen gas reacts with 4 moles of carbon to produce 1 mole of butane
This implies that
0.79 moles react with 1.12 moles of carbon to produce x moles of butane
Therefore
[tex]x = \frac{0.79}{5}[/tex]
[tex]x = 0.158 \ moles[/tex]
Now since this reaction is carried out at standard temperature and pressure the volume of butane produced will be
[tex]V_b = 0.158 * 22.4[/tex]
[tex]V_b = 3.54 \ L[/tex]
