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\(a.C_2H_5OH+3O_2-^{t^o}\rightarrow2CO_2+3H_2O\\ n_{C_2H_5OH}=0,3\left(mol\right)\\ n_{CO_2}=2n_{C_2H_5OH}=0,6\left(mol\right)\\ \Rightarrow V_{CO_2}=0,6.22,4=13,44\left(l\right)\\ b.n_{O_2}=3n_{C_2H_5OH}=0,6\left(mol\right)\\ MàV_{O_2}=\dfrac{1}{5}V_{kk}\\ \Rightarrow V_{kk}=V_{O_2}.5=0,6.22,4.5=67,2\left(l\right)\\ c.n_{NaOH}=0,9\left(mol\right)\\ Tacó:\dfrac{n_{NaOH}}{n_{CO_2}}=\dfrac{0,9}{0,6}=1,5\\ \Rightarrow Tạora2muốiNaHCO_3vàNa_2CO_3\\ Đặt:n_{NaHCO_3}=x\left(mol\right);n_{Na_2CO_3}=y\left(mol\right)\\ \Rightarrow\left\{{}\begin{matrix}x+y=0,6\left(BTnguyento\left(C\right)\right)\\x+2y=0,9\left(BTnguyento\left(Na\right)\right)\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x=0,3\\y=0,3\end{matrix}\right.\\ \Rightarrow m_{muối}=0,3.84+0,3.106=57\left(g\right)\)
a, PT: \(C_2H_4+3O_2\underrightarrow{t^o}2CO_2+2H_2O\)
Ta có: \(n_{C_2H_4}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)
Theo PT: \(n_{O_2}=3n_{C_2H_4}=0,75\left(mol\right)\)
\(\Rightarrow V_{O_2}=0,75.2,24=16,8\left(l\right)\)
\(\Rightarrow V_{kk}=16,8.5=84\left(l\right)\)
b, Theo PT: \(n_{CO_2}=2n_{C_2H_4}=0,5\left(mol\right)\)
PT: \(CO_2+Ca\left(OH\right)_2\rightarrow CaCO_{3\downarrow}+H_2O\)
Theo PT: \(n_{Ca\left(OH\right)_2}=n_{CaCO_3}=n_{CO_2}=0,5\left(mol\right)\)
\(\Rightarrow m_{\downarrow}=m_{CaCO_3}=0,5.100=50\left(g\right)\)
\(m_{Ca\left(OH\right)_2}=0,5.74=37\left(g\right)\)
\(\Rightarrow m_{ddCa\left(OH\right)_2}=\dfrac{37.100}{2}=1850\left(g\right)\)
Bạn tham khảo nhé!
PTHH: \(C_2H_5OH+3O_2\xrightarrow[]{t^o}2CO_2+3H_2O\)
Ta có: \(n_{C_2H_5OH}=\dfrac{6,4}{46}=\dfrac{16}{115}\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}n_{O_2}=\dfrac{48}{115}\left(mol\right)\\n_{CO_2}=\dfrac{32}{112}\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}V_{O_2}=\dfrac{48}{115}\cdot22,4\approx9,35\left(l\right)\\V_{CO_2}=\dfrac{32}{112}\cdot22,4\approx6,23\left(l\right)\end{matrix}\right.\)
\(n_{C_2H_2}=\dfrac{8,96}{22,4}=0,4mol\)
\(2C_2H_2+5O_2\rightarrow\left(t^o\right)4CO_2+2H_2O\)
0,4 1 0,8 ( mol )
\(V_{CO_2}=0,8.22,4=17,92l\)
\(V_{kk}=V_{O_2}.5=1.22,4.5=112l\)
\(n_{C_2H_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\\ PTHH:2C_2H_2+5O_2\underrightarrow{t^o}4CO_2+2H_2O\\ Mol:0,4\rightarrow1\rightarrow0,8\\ \rightarrow\left\{{}\begin{matrix}V_{CO_2}=0,8.22,4=17,92\left(l\right)\\V_{kk}=1.5.22,4=112\left(l\right)\end{matrix}\right.\)
a)
\(C + O_2 \xrightarrow{t^o} CO_2\)
\(b)\\ n_{O_2} = n_C = \dfrac{2,4}{12}=0,2(mol)\\ \Rightarrow V_{O_2} = 0,2.22,4 = 4,48(lít)\\ \Rightarrow V_{không\ khí} = 5V_{O_2} =4,48.5 = 22,4(lít)\)
a) PTHH : \(C+O_2-t^o-CO_2\)
b) \(n_C=\frac{2,4}{12}=0,2\left(mol\right)\)
Theo pthh : \(n_{O_2}=n_C=0,2\left(mol\right)\)
=> \(V_{O_2}=0,2\cdot22,4=4,48\left(l\right)\)
=> \(V_{kk}=\frac{4,48}{20}\cdot100=22,4\left(l\right)\)
a, n\(C_2H_6O\)= \(\frac{6,9}{46}\)=0,15mol
pt : C2H6O + 3O2 → 2CO2 + 3 H2O
(mol) 0,15mol → 0,45mol→0,3mol
m\(CO_2\) = 0,3 . 44 = 13,2 G
b, Vkk = \(\frac{20}{100}\) : (0,45 . 22,4 ) = 50,4l