How many grams of Cl are in a 38.0-g sample of CFCl3

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IrfanEinstein IrfanEinstein · Answer 1 · 2019-12-06T06:22:07+01:00

Answer:

38.0 g CF2Cl2) / (120.9135 g CF2Cl2/mol) x (2 mol Cl / 1 mol CF2Cl2) x (35.4532 g Cl/mol) =

22.3 g Cl in CF2Cl2

(38.0 g CFCl3) / (137.3681 g CFCl3/mol) x (3 mol Cl / 1 mol CFCl3) x (35.4532 g Cl/mol) =

29.4 g Cl in CFCl3

(38.0 g C2F3Cl3) / (187.3756 g C2F3Cl3/mol) x (3 mol Cl / 1 mol C2F3Cl3) x (35.4532 g Cl/mol) =

21.6 g Cl in C2F3Cl3

(38.0 g CF3Cl) / (104.4589 g CF3Cl/mol) x (1 mol Cl / 1 mol CF3Cl) x (35.4532 g Cl/mol) =

12.9 g Cl in CF3Cl

anamta123ahmed anamta123ahmed · Answer 2 · 2022-07-27T08:48:24+02:00

12.9 g Cl in CF3Cl

if 38.0 g CF2Cl2) / (120.9135 g CF2Cl2/mol) x (2 mol Cl / 1 mol CF2Cl2) x (35.4532 g Cl/mol) = 22.3 g Cl in CF2Cl2

then (38.0 g CFCl3) / (137.3681 g CFCl3/mol) x (3 mol Cl / 1 mol CFCl3) x (35.4532 g Cl/mol) = 29.4 g Cl in CFCl3

and then (38.0 g C2F3Cl3) / (187.3756 g C2F3Cl3/mol) x (3 mol Cl / 1 mol C2F3Cl3) x (35.4532 g Cl/mol) =21.6 g Cl in C2F3Cl3

finally the result is-(38.0 g CF3Cl) / (104.4589 g CF3Cl/mol) x (1 mol Cl / 1 mol CF3Cl) x (35.4532 g Cl/mol) =12.9 g Cl in CF3Cl

By following these steps we can calculate the no of grams of a given atom in the sample of a given element.

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