Suppose two chemical reactions are linked together in a way that the O2 produced inthe first reaction goes on to react completely with Mg to form MgO in the second reaction. Reaction one: 2 KClO3 3 O2→ + 2 KClReaction two: 2 Mg + O2 2 MgO→If you start with 4 moles of KClO3, how many moles of MgO could eventually form?

A) 2 moles
B) 4 moles
C) 6 moles
D) 12 moles
E) none of the above

12 moles of MgO could eventually form.

Explanation:

In the reaction:

2KClO₃ → 3O₂ + 2KCl

2 moles of KClO₃ produce 3 moles of O₂. 4 moles of KClO₃ produce:

4 mol KClO₃ × (3 mol O₂ / 2 mol KClO₃) = 6 mol O₂

In the second reaction:

2Mg + O₂ → 2MgO

1 mole of O₂ produce 2 moles of MgO. Thus, 6 mol of O₂ produce:

6 mol O₂ × (2 mol MgO  / 1 mol O₂) = 12mol MgO

Thus, 12 moles of MgO could eventually form.

the volume increases

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