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JEE Main & Advanced JEE Main Paper (Held On 09-Jan-2019 Evening)

  • question_answer
    For the following reaction, the mass of water produced from 4.45 g of \[{{C}_{57}}{{H}_{110}}{{O}_{6}}\] is: \[2{{C}_{57}}{{H}_{110}}{{O}_{6}}(s)+163{{O}_{2}}(g)\to 114C{{O}_{2}}(g)+110{{H}_{2}}O(\ell )\] [JEE Main Online Paper (Held On 09-Jan-2019 Evening]

    A) 445 g                                       

    B) 490 g

    C) 495g                

    D) 890 g

    Correct Answer: C

    Solution :

    Moles of \[{{C}_{57}}{{H}_{10}}{{O}_{6}}=\frac{445}{890}=0.5\] \[\frac{{{n}_{{{C}_{57}}{{H}_{110}}{{O}_{6}}}}}{2}=\frac{{{n}_{{{H}_{2}}O}}}{110}\] \[{{n}_{{{H}_{2}}O}}\,\,=55\,\,\times \,\,0.5\,\,\times \,\,18=495\,gm\]


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