The combustion of propane (C3H8) produces CO2 and H2O: C3H8 (g) + 5O2 (g) → 3CO2 (g) + 4H2O (g) The reaction of 1.0 mol of O2 will produce mol of H2O.

0.8 moles of H2O

Explanation:

C3H8 (g) + 502 (g) → 3C02 (g) + 4H2O(g)

Propane + Oxygen → Carbon dioxide + water

In the reaction above,

5 moles of O2 produce 4 moles of H2O

1.0 mol of O2 will produce Z moles of H2O

To get the value of Z, cross multiply

Z moles x 5 moles = 4 moles x 1.0 moles 5Z = 4

Z = 4/5

Z = 0.8

Thus, 1.0 moles of O2 will produce 0.8 moles of H2O

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