Question

For the following reaction, what volume of O2 would be required to react with 4.8 L of NO, measured at the same temperature and pressure? 2NO(g)+O2(g)-->2NO2(g)

Answer 1

2NO(g)      +    O2(g)    ------------------> 2NO2(g

2 x 22.4 L       22.4 L                           2 x 22.4 L

22.4 L of O2   needed ------------------> for 44.8 L of NO

   ??                           ------------------ > 4.8 L of NO

volume of O2 = 4.8 x 22.4 / 44.8

                    = 2.4 L

volume of O2 required = 2.4 L