At the candy store, sophie filled a bag with 2 2/3 kilograms of candy. 1/4 of the weight of the candy was from chocolate covered pretzels. how much did the chocolate covered pretzels in sophie’s bag weigh?

The answer is:  " ⅔ kg" ; or, write as:  " 0.6667 kg " .
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Explanation:
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   Basically, we can reduce the question to:
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          What is " ¼" of "2 ⅔ kg"  ?
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              →  (¼ * 2 ⅔)  kg ;  simplify the amount in "parentheses" ; & keep the units of "kg" (kilograms) ; 
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Write " 2 ⅔ " as an "improper fraction" ;  

→  To do so, multiply the "3" in the "denominator" (from the fraction, " ⅔ ") ; 
  by the "2" (from the whole number).   

      →  3 * 2 = 6 .
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→ Then, take this value; in our case, "6" ;  and the number "2" (from the "2" in the "numerator" (from the fraction portion, " ⅔ ").

      → 6 + 2 = 8 .  
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→ Then, take this value, and write is as the "numerator" over a "new fraction" ; this "new fraction" is the:  "improper fraction" . The "denominator" of this "new fraction" (i.e. "improper fraction") ; is the number from the "denominator" of the fraction portion of our [originally written "mixed number" ; which in our case is:  "3" (from the fraction portion, " ⅔ ") .
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→  So, our mixed number, " 2 ⅔ " , can be written in the "improper fraction form" ; which is:  " ⁸/₃ " .
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→  Put simply:  " 2 ⅔  = [ (3*2) + 2 ] / 3 = (6 + 2) / 3 = 8/3 ; or, " ⁸/₃ " .
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→ We have:   "(¼ * 2 ⅔)  kg " .

→ Replace the:  " 2 ⅔ " value with:  " ⁸/₃ " ;  to more easily simplify the expression in "parentheses" ; & to more easily solve the problem ; 
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 →  "(¼ * 2 ⅔) =   * " ;
                         
                         =    ?  ;
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→ We have:  " *  " ;

                  → The "4" can be 'canceled out' to a "1" ;
         and the "8" can be 'canceled out', and changed to a "2" ;
                             
                           →  since:  "(8÷4=2)" ; and since: "(4÷4=1)" .
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So now; we have:  " *  " ;
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  →  The " " can be eliminated; 

    → since:  "(1÷1=1)" ;
    → {Note that any value, divided by "1", is equal that same value ;
& that "any non-zero value", divided by "that same value", is equal to "1" .
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 → As such:  " "  =  {1÷1} = 1 ;
  & this value, which is "1",  multiplied by [the other expression], will be equal to that other expression;
since:
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→ [the other expression, which is a non-zero value] , multiplied by "1", is equal to [the SAME other expression, which is a non-zero value].
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→ As such, we can completely eliminate the:  " " ;

   and we are left with:  " ⅔ " .  The units are "kg" .

So; the answer is:  " ⅔ kg" ;  

Note:  " ⅔ " = 2÷3 = 0.666666666666667 ; 
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So; the answer is:  " ⅔ kg" ; or, write as:  0.6667 kg .
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