C.
If 150 grams of oxygen are produced, how many moles of sodium chlorate decomposed

3.13 moles of sodium chlorate are decomposed.

Explanation:

Given data:

Mass of oxygen produced = 150 g

Number of moles of sodium chlorate decomposed = ?

Solution:

Chemical equation:

2NaClO₃   →   2NaCl + 3O₂

Number of moles of oxygen:

Number of moles = mass/molar mass

Number of moles = 150 g/ 32 g/mol

Number of moles = 4.7 mol

Now we will compare the moles of oxygen with sodium chlorate.

                     O₂           :           NaClO₃

                      3             :              2

                    4.7            :          2/3×4.7 = 3.13 mol

3.13 moles of sodium chlorate are decomposed.

when the volume of gas is   changed from   717.73 ml   to 852 ml   the temperature will   change   from 315   c to 452   c

    calculation

this is calculated using the charles law formula

that is,   v1/t1 = v2/t2

          v1=?

        t1 = 315 c   into   kelvin = 315 +273 = 588 k

        v2 = 852   ml

        t2 = 425c + 273 =698 k

make v1   the subject of the formula ( by   multiplying both side by t1)

  v1=( v2 xt1) / t2

v1= (852 ml   x 588 k)/ 698 k= 717.73 ml

is there any options? maybe im just not seeing them but i need the options to

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