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xz_007 [3.2K]
2 years ago
7

The industrial preparation of ozone (03) uses oxygen gas (O2), as shown in the equation below. 302(8) electricity 203(g) A If 96

.0 kg of O2 gas actually yields 11.5 kg of 03. what is the percent yield in this reaction? A. 3.10% B. 8.35% C. 12.0% D. 18.0% ​
Chemistry
1 answer:
Paul [167]2 years ago
5 0

The percent yield of the industrial preparation of ozone is the 12%.

<h3>How do we calculate the percent yield?</h3>

Percent yield of any reaction will be calculated by using the below equation:

% yield = (Actual yield / Theoretical yield) × 100%

Given chemical reaction is:

3O₂(g) → 2O₃(g)

Moles of oxygen will be calculated as:

n = W/M, where

  • W = given mass = 96kg = 96,000g
  • M = molar mass = 32g/mol

n = 96,000/32 = 3000mol

From the stoichiometry of the reaction it is clear that:

3 moles of oxygen = produces 2 moles of ozone

3000 moles of oxygen = produces 2/3(3000)=2000 moles of ozone

Mass of 2000 moles of ozone = (2000mol)(48g/mol) = 96,000g = 96kg

Actual yield of ozone = 11.5kg (given)

Now percent yield will be calculated as:

% yield = (11.5 / 96) × 100% = 11.97% = 12%

Hence required percent yield is 12%.

To know more about percent yield, visit the below link:

brainly.com/question/20349874

#SPJ1

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What is the maximum mass of zinc oxide that can be produced by the reaction of 42.900000000000006g of zinc sulfide and 44.187000
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Now we will compare the moles of oxygen and zinc sulfide with zinc oxide.

                             ZnS        :       ZnO

                                2          :          2

                              0.44       :         0.44

                              O₂          :          ZnO

                              3           :            2

                              1.4          :           2/3×1.4 =0.93

The number of moles of zinc oxide produced by ZnS are less so it will limiting reactant.

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Mass = number of moles / molar mass

Mass = 0.44 mol / 81.38 g/mol

Mass = 0.00541 g

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Answer:   " 0.69 g / mL "

____________________________________________

Explanation:  

______________________________________________

The "pre-lab question" given is:

_____________________________________________

"The volume of an unknown liquid is 15 ml, and the mass of the liquid and the graduated cylinder it is in is 55.2g.   If the mass of the graduated cylinder is 44.8g , what is the density of the unknown liquid? " .

____________________________________________

Note:  The volume of the unknown liquid is 15 mL ;  regardless of whether or not the "unknown liquid" is in the graduated cylinder.

The density of the unknown liquid is measured in:  "g / mL " ;  

  that is, "grams per mL" .

____________________________________________

Note:  "D = m / V " ;   that is;  "Density  = mass/ volume" ;

                                  that is:   ["Density = mass per 'unit volume' ]" .

____________________________________________

So;  to find the density, "D" , of the "unknown liquid" ;  we would have to find the "mass" of the "unknown liquid" by "subtracting" :

       the "known mass of liquid when the liquid is not in the cylinder";  that is:  " 44.8g" ;  From:

       the "known combined value of the: 'mass of the liquid PLUS the mass of the cylinder" ;  that is:  "55.2g" ;

→     " 55.2g - 44.8g = 10.4g " .

___________________________________________

So:   " D =  m / V " ; (55.2g - 44.8 g) / 15 mL  " ;

              =  (55.2g - 44.8 g) / 15 mL  " ;

              =  (10.4g) / 15mL  ;

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Note that the "Density" =  mass per "unit volume" ;

→ So:  D = m / V ;  in units of:  " g / mL " (grams per millileter) ;

            = (10.4g) / 15mL  ;

             = [ (10.4) / 15 }  g/ mL ;  

            =  0.693333333333333.... g / mL  ;

       →  We round to "2 (two) significant figures" ;

       → since we have:  "10.4 g / 15 mL " ;

             and 15 mL is considered a measured/experimental value;

             and:  10.4g is considered a measured/experimental value;

→  so, the least precise value;  15 mL (has only 2 (two) significant figures ;

      compared to other value:  10.4g (which has 3 (three) significant figures;

→  so we shall round off to 2 (two) significant figures:

             =  0.69 g / mL

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              →  which is our answer:  " 0.69 g / mL " .

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3 years ago
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