\(\begin{array}{l} Mhh = 11 \times 2 = 22g/mol\\ nhh = \dfrac{{6,72}}{{22,4}} = 0,3\,mol\\ hh:{H_2}(a\,mol),{O_2}(b\,mol)\\ a + b = 0,3\\ 2a + 32b = 0,3 \times 22\\ = > a = 0,1;b = 0,2\\ \% V{H_2} = \dfrac{{0,1}}{{0,3}} \times 100\% = 33,33\% \\ \% V{O_2} = 100 - 33,33 = 66,67\% \\ 2{H_2} + {O_2} \to 2{H_2}O\\ n{H_2}O = n{H_2} = 0,1\,mol\\ n{O_2} = 0,2 - \dfrac{{0,1}}{2} = 0,15\,mol\\ m{H_2}O = 0,1 \times 18 = 1,8g\\ m{O_2} = 0,15 \times 32 = 4,8g \end{array}\)