cần bao nhiêu ml rượu 40 độ để phản ứng hoàn toàn 46g Na ( ai giúp bai này với đang cần gấp )
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\(V_{C_2H_5OH}=\dfrac{100.40}{100}=40\left(ml\right)\\ \rightarrow m_{C_2H_5OH}=40.0,8=32\left(g\right)\\ \rightarrow n_{C_2H_5OH}=\dfrac{32}{46}=\dfrac{16}{23}\left(mol\right)\)
PTHH: \(C_2H_5OH+Na\rightarrow C_2H_5ONa+\dfrac{1}{2}H_2\uparrow\)
\(\dfrac{16}{23}\)----------------------------------->\(\dfrac{8}{23}\)
\(\rightarrow V_{H_2}=\dfrac{8}{23}.22,4=\dfrac{896}{115}\left(l\right)\)
\(a,n_{CH_3COOH}=\dfrac{120.20}{100}=24\left(g\right)\\ \rightarrow n_{CH_3COOH}=\dfrac{24}{60}=0,4\left(mol\right)\)
PTHH: \(2CH_3COOH+Na_2CO_3\rightarrow2CH_3COONa+CO_2\uparrow+H_2O\)
0,4----------->0,2-------------->0,4-------------->0,2
\(\rightarrow m_{ddNa_2CO_3}=\dfrac{0,2.106}{10\%}=212\left(g\right)\)
\(\rightarrow m_{ddA}=212+120-0,2.44=323,2\left(g\right)\\ \rightarrow C\%_{CH_3COONa}=\dfrac{0,4.82}{323,2}.100\%=10,15\%\)
b, PTHH: \(C_2H_5OH+O_2\underrightarrow{\text{men giấm}}CH_3COOH+H_2O\)
0,4<------------------------0,4
\(\rightarrow V_{ddC_2H_5OH}=\dfrac{0,4.46.100}{0,8.46}=50\left(ml\right)\)
a) C2H5OH + 3O2 --to--> 2CO2 + 3H2O
b) \(n_{C_2H_5OH}=\dfrac{46}{46}=1\left(mol\right)\)
PTHH: C2H5OH + 3O2 --to--> 2CO2 + 3H2O
1----->3----------->2------->3
=> VO2 = 22,4.3 = 67,2 (l)
c) mH2O = 3.18 = 54 (g)
d) VCO2 = 2.22,4 = 44,8 (l)
a) \(PTHH:2SO_2+O_2\xrightarrow[V_2O_5]{450^oC}2SO_3\)
\(n_{SO_2}=\dfrac{32}{64}=0,5\left(mol\right)\\ n_{O_2}=\dfrac{10}{32}=0,3125\left(mol\right)\)
Lập tỉ lệ: \(\dfrac{n_{SO_2}}{2}< \dfrac{n_{O_2}}{1}\left(\dfrac{0,5}{2}< 0,3125\right)\)
=> SO2 hết O2 dư
Theo pt: \(n_{O_2\left(pư\right)}=\dfrac{n_{SO_2}.2}{3}=\dfrac{0,5.1}{2}=0,25\left(mol\right)\)
\(n_{O_2\left(dư\right)}=0,3125-0,25=0,0625\left(mol\right)\\ m_{O_2}=0,0625.32=2\left(g\right)\)
c) Theo pt, ta có:\(n_{SO_3}=n_{SO_2}=0,5\left(mol\right)\)
\(m_{SO_3}=0,5.80=40\left(g\right)\)
a) nNaOH = 0,6.1 = 0,6 (mol)
PTHH: NaOH + CH3COOH --> CH3COONa + H2O
0,6----->0,6
=> mCH3COOH = 0,6.60 = 36 (g)
=> mC2H5OH = 45,2 - 36 = 9,2 (g)
b) \(n_{C_2H_5OH}=\dfrac{9,2}{46}=0,2\left(mol\right)\)
PTHH: 2CH3COOH + 2Na --> 2CH3COONa + H2
0,6---------------------------->0,3
2C2H5OH + 2Na --> 2C2H5ONa + H2
0,2--------------------------->0,1
=> V = (0,3 + 0,1).22,4 = 8,96 (l)
Câu 2:
PTHH: \(Na+C_2H_5OH\rightarrow C_2H_5ONa+\dfrac{1}{2}H_2\uparrow\)
Ta có: \(n_{C_2H_5OH}=\dfrac{40}{46}=\dfrac{20}{23}\left(mol\right)=n_{Na}\)
\(\Rightarrow m_{Na}=\dfrac{20}{23}\cdot23=20\left(g\right)\)
a)\(n_{O_2}=\dfrac{10,08}{22,4}=0,45mol\)
\(2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\)
0,9 0,45
\(m_{KMnO_4}=0,9\cdot158=142,2g\)
b)\(m_{K_2MnO_4}=0,45\cdot197=88,65g\)
c)\(2Fe+O_2\underrightarrow{t^o}2FeO\)
0,9 0,45
\(m_{Fe}=0,9\cdot56=50,4g\)
a, \(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\)
PT: \(Mg+2HCl\rightarrow MgCl_2+H_2\)
\(Mg+H_2SO_4\rightarrow MgSO_4+H_2\)
Gọi: Vhh axit = a (l)
\(\Rightarrow\left\{{}\begin{matrix}n_{HCl}=1,5a\left(mol\right)\\n_{H_2SO_4}=0,5a\left(mol\right)\end{matrix}\right.\)
Theo PT: \(n_{Mg}=\dfrac{1}{2}n_{HCl}+n_{H_2SO_4}\) \(\Rightarrow0,2=\dfrac{1}{2}.1,5a+0,5a\)
⇒ a = 0,16 (l) = 160 (ml)
b, Theo PT: \(n_{H_2}=n_{Mg}=0,2\left(mol\right)\Rightarrow V_{H_2}=0,2.22,4=4,48\left(l\right)\)