\(n_{H_2} = \dfrac{11,2}{22,4} = 0,5(mol)\\ Cr_2O_3 + 3H_2 \xrightarrow{t^o} 2Cr + 3H_2O\\ n_{Cr} = \dfrac{2}{3}n_{H_2} = \dfrac{1}{3}(mol)\\ m_{Cr}= \dfrac{1}{3}.52 = 17,33(gam)\\ n_{H_2O} = n_{H_2} =0,5(mol) \\ \Rightarrow m_{H_2O} = 0,5.18 = 9(gam)\)