\(n_{O_2\left(tt\right)}=\dfrac{40000}{32}=1250\left(mol\right)\)
\(\Rightarrow n_{o_2\left(PTHH\right)}=\dfrac{1250\cdot100\%}{85\%}=\dfrac{25000}{17}\left(mol\right)\\ 2H_2O\underrightarrow{dp}2H_2+O_2\)
Theo PTHH => nH2O=2nO2=2*25000/17(mol)
\(m_{H_2O\left(can\right)}=\dfrac{25000}{17}\cdot2\cdot18\approx52941,18\left(g\right)\)