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Câu hỏi: Lên men 45 gam glucozơ để điều chế ancol etylic (hiệu suất phản ứng lên men là 80%, thu được V lít khí CO2 (đktc). Giá trị của V là
A. 5,6.
B. 11,2
C. 4,48.
D. 8,96.
A. 5,6.
B. 11,2
C. 4,48.
D. 8,96.
Phương pháp:
- Tính theo PTHH: ${{C}_{6}}{{\text{H}}_{12}}{{\text{O}}_{6}}\xrightarrow{menruou}2{{\text{C}}_{2}}{{\text{H}}_{5}}\text{OH}+2\text{C}{{\text{O}}_{2}}$
- Công thức tính hiệu suất: ${{\text{H}}^{0}}\%=\dfrac{{{n}_{\text{C}{{\text{O}}_{2}}(\text{TT})}}}{{{\text{n}}_{\text{C}{{\text{O}}_{2}}(L\text{T})}}}.100\%$
Hướng dẫn giải:
${{n}_{{{C}_{6}}{{H}_{l2}}{{C}_{6}}}}=\dfrac{45}{180}=0,25(\text{mol})$
PTHH ${{\text{C}}_{6}}{{\text{H}}_{12}}{{\text{O}}_{6}}\xrightarrow{menruou}\text{2}{{\text{C}}_{2}}{{\text{H}}_{5}}\text{OH}+2\text{C}{{\text{O}}_{2}}$
Theo PTHH $\Rightarrow {{n}_{\text{C}{{\text{O}}_{2}}\left( \text{LT} \right)}}=2{{n}_{{{\text{C}}_{6}}{{\text{H}}_{12}}{{O}_{6}}}}=0,5(\text{mol})$
Do $\text{H}=80\%\Rightarrow {{\text{n}}_{C{{O}_{2}}\left( \text{rr} \right)}}=0,5\cdot \dfrac{80}{100}=0,4(\text{mol})$
$\Rightarrow \text{V}=0,4.22,4=8,96$ lít
- Tính theo PTHH: ${{C}_{6}}{{\text{H}}_{12}}{{\text{O}}_{6}}\xrightarrow{menruou}2{{\text{C}}_{2}}{{\text{H}}_{5}}\text{OH}+2\text{C}{{\text{O}}_{2}}$
- Công thức tính hiệu suất: ${{\text{H}}^{0}}\%=\dfrac{{{n}_{\text{C}{{\text{O}}_{2}}(\text{TT})}}}{{{\text{n}}_{\text{C}{{\text{O}}_{2}}(L\text{T})}}}.100\%$
Hướng dẫn giải:
${{n}_{{{C}_{6}}{{H}_{l2}}{{C}_{6}}}}=\dfrac{45}{180}=0,25(\text{mol})$
PTHH ${{\text{C}}_{6}}{{\text{H}}_{12}}{{\text{O}}_{6}}\xrightarrow{menruou}\text{2}{{\text{C}}_{2}}{{\text{H}}_{5}}\text{OH}+2\text{C}{{\text{O}}_{2}}$
Theo PTHH $\Rightarrow {{n}_{\text{C}{{\text{O}}_{2}}\left( \text{LT} \right)}}=2{{n}_{{{\text{C}}_{6}}{{\text{H}}_{12}}{{O}_{6}}}}=0,5(\text{mol})$
Do $\text{H}=80\%\Rightarrow {{\text{n}}_{C{{O}_{2}}\left( \text{rr} \right)}}=0,5\cdot \dfrac{80}{100}=0,4(\text{mol})$
$\Rightarrow \text{V}=0,4.22,4=8,96$ lít
Đáp án D.