Giải bài 8 trang 48 - SGK môn Hóa học lớp 11 nâng cao

${{V}_{KK}}=841,7{{m}^{3}}\Rightarrow {{V}_{{{N}_{2}}}}\approx 656,7{{m}^{3}};\,{{V}_{{{O}_{2}}}}\approx 177{{m}^{3}}$

Tỉ lệ ${{N}_{2}}:{{H}_{2}}=1:3\Rightarrow {{V}_{{{H}_{2}}}}\approx 1970{{m}^{3}}$

$\begin{align} & C{{H}_{4}}+2{{H}_{2}}O\to C{{O}_{2}}+4{{H}_{2}}\,\,\,\,\,\,\, \\ & 1{{m}^{3}}\,\,\,\,\,\,\,\,2{{m}^{3}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,4{{m}^{3}} \\ & x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,y\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1970{{m}^{3}} \\ & \Rightarrow {{V}_{{{H}_{2}}O}}=y=985{{m}^{3}} \\ & x=492,5{{m}^{3}} \\ & \\ \end{align}$

$\begin{align} & C{{H}_{4}}+2{{O}_{2}}\to C{{O}_{2}}+{{H}_{2}}O \\ & 1{{m}^{3}}\,\,\,\,\,\,\,\,\,2{{m}^{3}} \\ & z\,\,\,\,\,\,\,\,\,\,\,\,\,\,177{{m}^{3}} \\ & \Rightarrow z=88,5{{m}^{3}} \\ & \Rightarrow {{V}_{C{{H}_{4}}}}=x+z=581{{m}^{3}} \\ \end{align}$

Ghi nhớ:

Tỉ lệ thể tích của nitơ và hiđro trong phản ứng điều chế amoniac là 1:3