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EAMCET Medical EAMCET Medical Solved Paper-2008

  • question_answer
    40 g of a sample of carbon on combustion left 10% of it unreacted. The volume of oxygen required at STP for this combustion reaction is

    A)  22.4 L                                   

    B)  67.2 L

    C)  11.2 L                                   

    D)  44.8 L

    Correct Answer: B

    Solution :

                     \[\underset{12\,g}{\mathop{C}}\,+\underset{22.4\,L}{\mathop{{{O}_{2}}}}\,\xrightarrow{{}}C{{O}_{2}}\] The amount of carbon unreacted \[=40\times \frac{10}{100}g\] \[=4\,g\,\] So, the amount of carbon reacted \[=(40-4)g\] \[=36\,g\] \[\because \] At STP, for the combustion of 12 g of C, oxygen required is = 22.4 L                    \[\therefore \]For the combustion of 36 g of C, oxygen    required will be \[=\frac{22.4}{12}\times 36L=67.2\,L\]


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