The combustion of methanol (CH4O) is described by this balanced chemical equation: 2CH4O(g) + 3O2(g) → 4H2O(g) + 2CO2(g)

If the reaction is conducted and maintained at standard temperature and pressure (STP), how does the volume change when all of the reactants combine to form products?

volume stays the same

volume increases by a factor of 2

volume decreases by a factor of 5/6

volume increases by a factor of 6/5

im not exactly sure but i think its b!

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

in the analysis of systems that involve fluid flow, we frequently encounter the combination of properties u and pv.

c

explanation: