\(7-D\\ 8-C\\ 9-B\\ 10-C\\ n_{H_2}=0,15\left(mol\right);n_{O_2}=0,05\left(mol\right)\\ 2H_2+O_2-^{t^o}\rightarrow2H_2O\\ LTL:\dfrac{0,15}{2}>\dfrac{0,05}{1}\\ \Rightarrow H_2dư\\ n_{H_2O}=2n_{O_2}=0,1\left(mol\right)\\ \Rightarrow m_{H_2O}=1,8\left(g\right)\\ 11-A\\ 12-D\)