Consider the following reaction involved in the preparation of teflon polymer (−CF2−CF2−)n XeF2+(−CH2−CH2−)n→(−CF2−CF2−)n+HF+XeFa−. Determine the moles of XeF5 required for preparation of 100g Teflon.
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Solution
XeF2+(CH2−CH2)n→(CF2−CF2)n+XeFa
for one (CF2−CF2) the mass=100gm
formation one teflon, for which n=1
2XeF2+(CH2−CH2)(CF2−C2)1+2Xe
2 moles of XeF2 produce 1 mole of (CF2−CF2)1
whcih is 100gm teflon so, 2 moles of XeF2 is requried to produce 100gm of Teflon.