CHEMISTRY

JAMB 2006 - Question 40

Chemistry 2006 JAMB Past Questions - Question 40: In the reaction above, the mass of carbon(IV)oxide produced on burning 78 g of ethyne is

Choose the correct answers from the options given.
In the reaction above, the mass of carbon(IV)oxide produced on burning 78 g of ethyne is
A:
B:
C:
D:
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Correct Answer

A

Explanation

  

The balanced chemical equation for the combustion of ethyne (also known as acetylene) is:

\[ 2\text{C}_2\text{H}_2 + 5\text{O}_2 \rightarrow 4\text{CO}_2 + 2\text{H}_2\text{O} \]

This equation indicates that 2 moles of ethyne react with 5 moles of oxygen to produce 4 moles of carbon dioxide and 2 moles of water.

Now, let's calculate the molar mass of ethyne (C2H2):

\[ \text{Molar mass of C2H2} = (2 \times \text{atomic mass of C}) + (2 \times \text{atomic mass of H}) \]

\[ \text{Molar mass of C2H2} = (2 \times 12.01 \, \text{g/mol}) + (2 \times 1.01 \, \text{g/mol}) = 26.03 \, \text{g/mol} \]

Given that you have 78 g of ethyne, you can now calculate the moles of ethyne:

\[ \text{Moles of C2H2} = \frac{\text{Mass}}{\text{Molar mass}} \]

\[ \text{Moles of C2H2} = \frac{78 \, \text{g}}{26.03 \, \text{g/mol}} \approx 2.992 \, \text{mol} \]

Now, using the stoichiometry of the balanced equation, we can determine the moles of carbon dioxide produced. Since the ratio of moles of ethyne to moles of carbon dioxide is 2:4, or simplified, 1:2, the moles of carbon dioxide produced will be half of the moles of ethyne.

\[ \text{Moles of CO2} = \frac{1}{2} \times \text{Moles of C2H2} \]

\[ \text{Moles of CO2} = \frac{1}{2} \times 2.992 \, \text{mol} = 1.496 \, \text{mol} \]

Finally, to find the mass of carbon dioxide produced, use the molar mass of carbon dioxide (CO2):

\[ \text{Molar mass of CO2} = (\text{atomic mass of C} + 2 \times \text{atomic mass of O}) \]

\[ \text{Molar mass of CO2} = (12.01 \, \text{g/mol} + 2 \times 16.00 \, \text{g/mol}) = 44.01 \, \text{g/mol} \]

Now, calculate the mass of carbon dioxide:

\[ \text{Mass of CO2} = \text{Moles of CO2} \times \text{Molar mass of CO2} \]

\[ \text{Mass of CO2} = 1.496 \, \text{mol} \times 44.01 \, \text{g/mol} \approx 65.89 \, \text{g} \]

Therefore, the mass of carbon dioxide produced on burning 78 g of ethyne is approximately 65.89 g.

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