giúp e vs ạ cần gấp Tính số gam nước thu được khi cho 8,4 lít khí hidro tác dụng với khí oxi( các thể tích ở đtkc) *

PTHH

$
\begin{array}{l}
{\hspace{0.33em}{2}{H}_{2}\mathrm{{+}}{O}_{2}\mathrm{\rightarrow}{2}{H}_{2}{O}}\\
{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}{n}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}{n}}\\
{\hspace{0.33em}\hspace{0.33em}{nH}_{2}\mathrm{{=}}\frac{8\mathrm{,}4}{\mathrm{22}\mathrm{,}4}\mathrm{{=}}{0}{\mathrm{,}}{3}{7}{5}{\mathrm{(}}{mol}{\mathrm{)}}}\\
{\hspace{0.33em}\hspace{0.33em}{nH}_{2}{O}\mathrm{{=}}{0}{\mathrm{,}}{3}{7}{5}{\mathrm{(}}{mol}{\mathrm{)}}}\\
{\hspace{0.33em}\hspace{0.33em}{mH}_{2}{O}\mathrm{{=}}{0}{\mathrm{,}}{3}{7}{5}{\mathrm{.}}{\mathrm{(}}{2}\mathrm{{+}}{1}{6}{\mathrm{.}}{2}{\mathrm{)}}}\\
{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\mathrm{{=}}{\mathrm{12}}{\mathrm{,}}{\mathrm{75}}{\mathrm{(}}{g}{\mathrm{)}}}
\end{array}
$