11th Class Chemistry Some Basic Concepts of Chemistry

  • question_answer 12) Calculate the amount of carbon dioxide that could be produced when: (i) 1 mole of carbon is burnt in air. (ii) 1 mole of carbon is burnt in 16 g of dioxygen. (iii) 2 moles of carbon are burnt in 16 g of dioxygen.

    Answer:

    (i) \[\underset{1\,\text{mol}}{\mathop{C(s)}}\,+{{O}_{2}}(g)\to \underset{1\,\text{mol}}{\mathop{C{{O}_{2}}(g)}}\,\] Above reaction shows that 1 \[mol\] carbon, when burnt in excess air will give 1 \[mol\]\[C{{O}_{2}}.\] (ii) (1 \[mol\] carbon = 12 g carbon) and 16 g of dioxygen are allowed to react, according to the following reaction: \[\underset{\underset{1\times 12g}{\mathop{1\,\text{mol}}}\,}{\mathop{C(s)}}\,+\underset{\underset{1\times \,32g}{\mathop{1\,\text{mol}}}\,}{\mathop{{{O}_{2}}(g)}}\,\to \underset{\underset{1\times \,44g}{\mathop{1\text{mol}}}\,}{\mathop{C{{O}_{2}}(g)}}\,\] Case I: Let carbon is completely consumed in the reaction: 12 g carbon on complete consumption will give 44 g \[C{{O}_{2}}\] Case II: Let oxygen is completely consumed in the reaction: \[\because \] \[32g\]oxygen gives 44 g \[C{{O}_{2}}\] \[\therefore \] 16 g oxygen will give 22 g \[C{{O}_{2}}\] Since, oxygen on complete consumption gives least amount of product (\[C{{O}_{2}}\]) hence oxygen is limiting reactant and hence 22 g will be actually formed in the reaction. (iii) 2 \[mol\] carbon i.e., 24 g carbon is allowed to react with 16 g dioxygen. \[\underset{\underset{1\times \,12g}{\mathop{1\,\text{mol}}}\,}{\mathop{C(s)}}\,+\underset{\underset{1\,\times \,32g}{\mathop{1\,\text{mol}}}\,}{\mathop{{{O}_{2}}(g)}}\,\to \underset{\underset{1\times \,44g}{\mathop{1\,\text{mol}}}\,}{\mathop{C{{O}_{2}}(g)}}\,\] Case I: Let carbon is completely consumed in the reaction: \[\because \]12 g carbon gives 44 g \[C{{O}_{2}}\] \[\therefore \]24 g carbon will give 88 g \[C{{O}_{2}}\] Case II: Let oxygen is completely consumed in the reaction: \[\because \]32 g oxygen gives 44 g \[C{{O}_{2}}\] \[\therefore \]16 g oxygen will give 22 g \[C{{O}_{2}}\] Since, oxygen on complete consumption gives least amount of product (\[C{{O}_{2}}\]) thus, oxygen is limiting reagent and hence 22 g \[C{{O}_{2}}\] will be actually formed in the reaction.


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