Assuming all volume measurements are made at the same temperature and pressure, what volume of hydrogen gas is needed to react completely with 4.55 l of oxygen gas to produce water vapor?

9.1L

Explanation:

Firstly, to solve the problem, we need a balanced chemical equation.

2H2 + O2 ———> 2H2O

2 moles of hydrogen reacted one mole of oxygen. Now, we know that at STP, one mole of a gas occupies a volume of 22.4L, now let us get the number of moles in 4.55L of oxygen.

This means 4.55/22.4 = 0.203125 mole

Since in theory we have 2 moles of hydrogen reacting one mole of oxygen. Hence the number of moles of hydrogen we have is 0.203125 * 2 = 0.40625 mole

We now proceed to get the volume of hydrogen gas. Since 1 mole is 22.4L, then

0.40625 Is 0.40625 * 22.4 = 9.1L

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

i do not get it   at all