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) * 16/11/2021 Bởi Skylar 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) *
nH2 =0,375 ; nO2 =0,125 PTHH: O2+2H2→2H2O Theo PTHH, hidro dư nên số mol nước tạo thành là: nH2O=2nO2=2×0,125=0,25 mol. Số gam nước thu được là: mH2O=0,25×18=4,5 gam. Bình luận
PTHH $\begin{array}{l}{\hspace{0.33em}{2}{H}_{2}\mathrm{{+}}{O}_{2}\mathrm{\rightarrow}{2}{H}_{2}{O}}\\{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}{n}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}{n}}\\{\hspace{0.33em}\hspace{0.33em}{nH}_{2}\mathrm{{=}}\frac{8\mathrm{,}4}{\mathrm{22}\mathrm{,}4}\mathrm{{=}}{0}{\mathrm{,}}{3}{7}{5}{\mathrm{(}}{mol}{\mathrm{)}}}\\{\hspace{0.33em}\hspace{0.33em}{nH}_{2}{O}\mathrm{{=}}{0}{\mathrm{,}}{3}{7}{5}{\mathrm{(}}{mol}{\mathrm{)}}}\\{\hspace{0.33em}\hspace{0.33em}{mH}_{2}{O}\mathrm{{=}}{0}{\mathrm{,}}{3}{7}{5}{\mathrm{.}}{\mathrm{(}}{2}\mathrm{{+}}{1}{6}{\mathrm{.}}{2}{\mathrm{)}}}\\{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\mathrm{{=}}{\mathrm{12}}{\mathrm{,}}{\mathrm{75}}{\mathrm{(}}{g}{\mathrm{)}}}\end{array}$ Bình luận
nH2 =0,375 ; nO2 =0,125
PTHH: O2+2H2→2H2O
Theo PTHH, hidro dư nên số mol nước tạo thành là:
nH2O=2nO2=2×0,125=0,25 mol.
Số gam nước thu được là:
mH2O=0,25×18=4,5 gam.
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}
$