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JEE Main2012ChemistrySome Basic Concepts of ChemistryActual

When CO _ 2( ~g ) is passed over red hot coke it partially gets reduced to CO (g) . Upon passing 0.5 ~L of CO ₂(g) over red hot coke, the total volume of the gases increased to 700 ~mL . The composition of the gaseous mixture at STP is

Options

  1. ACO ₂=300 ~mL ; CO =400 ~mL
  2. BCO ₂=0.0 ~mL ; CO =700 ~mL
  3. CCO ₂=200 ~mL ; CO =500 ~mL
  4. DCO ₂=350 ~mL ; CO =350 ~mL

Correct answer

A. CO ₂=300 ~mL ; CO =400 ~mL

Step-by-step solution

CO ₂+ C 2 CO Stoichoimetry ratio is 1: 2 AT STP, P=1 ~atm , T=273 ~K , R=0.0821 Initial moles of CO ₂ ; n ( CO ₂ . initial )= P V R T = 1 0.5 0.0821 273 =0.022 ~mole In final mixture no. of moles; n ( CO ₂ / CO . mixture) = 1 0.7 0.0821 273 =0.031 Increase in volume is by =0.031-0.022=0.009 mole of gas Final no. of moles of CO i.e. n_ (CO final) aligned & n_ (CO final) =2 n_ ( CO ₂ initial ) -n_ ( CO ₂ . final) & =2 (0.022- n _ ( CO ₂ final ) . & .n_ (CO final) =0.044-2 n_ (CO final ) & Now, n_ (CO final) +n_ (CO f

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