Question

When $ 12.0g $ of carbon reacts with oxygen to form $ CO $ and $ C{O_2} $ at $ {25^o}C $ and constant pressure, $ 75.0Kcal $ of heat was liberated and no carbon remained. Calculate the mass of oxygen needed for it and the mole of $ CO $ and $ C{O_2} $ formed.
[ given that: $ {\Delta _f}HC{O_2} = - 94.05 $ and $ {\Delta _f}HCO = - 26.41kcal\;mo{l^{ - 1}} $ ]

Answer 1

Hint: To give the answer to the above question we need to first understand what we are provided in the question. By using the information provided and the laws of chemistry and stoichiometry we can find the amount of oxygen used.

Complete step by step answer:
First of all, we will write a balanced chemical reaction of formation of both carbon dioxide and carbon monoxide with carbon and oxygen as reactants.
Carbon monoxide:
 $ C + 1/2{O_2} \to CO $
It will take 1/2 $ {O_2} $ to form carbon monoxide.
Carbon dioxide:
 $ C + {O_2} \to C{O_2} $
It will take 1 $ {O_2} $ to form carbon dioxide.
Now let assume that we used $ x $ moles of carbon for carbon dioxide and $ y $ moles of carbon for carbon monoxide. And the total amount of carbon used is $ 12g $ which is one mole of carbon. Therefore the total moles of carbon is;
 $ \Rightarrow $ $ x + y = 1 ………………. (1) $
Now we are provided with the heat of formation of carbon dioxide and heat of formation of carbon monoxide;
Heat of formation of carbon dioxide: $ {\Delta _f}HC{O_2} = - 94.05 $
Heat of formation of carbon monoxide: $ {\Delta _f}HCO = - 26.41kcal\;mo{l^{ - 1}} $
and the total energy formed is $ 75.0Kcal $
 $ \Rightarrow $ $ (x*94.05) + (y*26.41) = 75.0Kcal $
 $ \Rightarrow $ $ 94.01x + 26.41y = 75.0Kcal $
Now substituting equation (1) in above equation we get
 $ \Rightarrow $ $ 94.01x + 26.41(1 - x) = 75.0Kcal $
 $ \Rightarrow $ $ 94.01x + 26.41 - 26.41x = 75.0Kcal $
 $ \Rightarrow $ $ 67.6x + 26.41 = 75.0Kcal $
 $ \Rightarrow $ $ 67.6x = 75.0 - 26.41Kcal $
 $ \Rightarrow $ $ 67.6x = 48.59Kcal $
 $ \Rightarrow $ $ x = 48.59/67.6Kcal $
 $ \Rightarrow $ $ x = 0.718Kcal $
The moles of carbon dioxide used is $ 0.718moles $
Now substitute the value of $ x $ in equation (1)
 $ \Rightarrow $ $ 0.718 + y = 1 $
 $ \Rightarrow $ $ y = 1 - 0.718 $
 $ \Rightarrow $ $ y = 0.282 $
The moles of carbon monoxide used is $ 0.282moles $
According to stoichiometry the same amount of oxygen is used, therefore the moles of oxygen used for carbon dioxide is $ 0.718moles $ and since ½ $ {O_2} $ is used in carbon monoxide the moles of oxygen will be half from that of $ 0.282moles $ that is $ 0.141moles $ . So the total moles of oxygen is
 $ \Rightarrow $ $ mole{s_{{O_2}}} = 0.718 + 0.141 $
 $ \Rightarrow $ $ mole{s_{{O_2}}} = 0.859moles $
Now the total amount of oxygen used is;
 $ \Rightarrow $ $ {m_{{O_2}}} = mole{s_{{O_2}}}*32 $ (32 is the molecular mass of oxygen molecule)
 $ \Rightarrow $ $ {m_{{O_2}}} = 0.859*32 $
 $ \Rightarrow $ $ {m_{{O_2}}} = 27.49g $
The total mass of oxygen used is $ 27.49g $

Note:
You can always use stoichiometry if a reaction is given and you are asked about the mass of one of the compounds, but always remember to balance the chemical reaction first before calculating any of the quantity.