JEE Main & Advanced Sample Paper JEE Main Sample Paper-6

  • question_answer
    When \[Fe{{S}_{2}}\] is burnt in air, it converts into \[F{{e}_{2}}{{S}_{3}}\]. The change in percentage by weight of iron in the process is (Fe = 56)

    A)  23% increase    

    B)  12% decrease

    C)  12% increase   

    D)  no change

    Correct Answer: A

    Solution :

    \[\underset{\text{Fe }\!\!%\!\!\text{ }\frac{\text{56}}{\text{120}}\text{ }\!\!\times\!\!\text{ 100=46}\text{.67}}{\mathop{\underset{\text{56+32 }\!\!\times\!\!\text{ 2=120}}{\mathop{\text{Fe}{{\text{S}}_{\text{2}}}}}\,}}\,\xrightarrow[{}]{{}}\underset{\frac{112}{\text{160}}\text{ }\!\!\times\!\!\text{ 100}\,\text{=}\,\text{70}}{\mathop{\underset{\text{56}\times \text{2+3 }\!\!\times\!\!\text{ 16}\,\text{=160}}{\mathop{\text{F}{{\text{e}}_{\text{2}}}{{O}_{3}}}}\,}}\,\] Percent increase = 70 - 46.67 = 23% (approx.)


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