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JEE Advanced2021ChemistrySome Basic Concepts of ChemistryActual

For the following reaction scheme, percentage yields are given along the arrow: x g and y g are mass of R and U , respectively. (Use: Molar mass (in g mol ⁻¹ ) of H , C and O as 1,12 and 16 , respectively) The value of x is ___.

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0

Step-by-step solution

Mg 2 C 3 → H 2 O Mg ( OH ) 2 + C 3 H 4 ( P ) Propyne   ( 4 g ) Now, one mole of propyne give, one mole of 2 -butyne. 3 moles of 2 -butyne give one mole of hexamethyl benzene. Hence, 40   g of propyne give 54   g to 2 -butyne 100 % but given that 75 % yeild and 4   g to propyne. The mass of 2 -butyne formed = 5 .4 × 75 100 = 4 .05   g 162   g to 2 -butyne give 162   g of hexamethyl benzene 100 %   yield But given that 40 % yield, and 4 . 05   g of 2 -butyne The

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