Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
\(Mg+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Mg+H_2\)
0,2 0,1 ( mol )
\(\left\{{}\begin{matrix}m_{CH_3COOH}=0,2.60=12\left(g\right)\\m_{C_2H_5OH}=16,6-12=4,6\left(g\right)\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\%m_{CH_3COOH}=\dfrac{12}{16,6}.100=72,29\%\\\%m_{C_2H_5OH}=100-72,29=27,71\%\end{matrix}\right.\)
\(n_{C_2H_5OH}=\dfrac{4,6}{46}=0,1\left(mol\right)\)
\(C_6H_{12}O_6\xrightarrow[men.rượu]{30^o-35^o}2C_2H_5OH+2CO_2\)
0,05 0,1 ( mol )
\(m_{dd_{C_6H_{12}O_6}}=\dfrac{0,05.180.100}{15}=60\left(g\right)\)
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)
Có lẽ đề hỏi %m mỗi chất bạn nhỉ?
Ta có: 46nC2H5OH + 60nCH3COOH = 21,2 (1)
PT: \(C_2H_5OH+Na\rightarrow C_2H_5ONa+\dfrac{1}{2}H_2\)
\(CH_3COOH+Na\rightarrow CH_3COONa+\dfrac{1}{2}H_2\)
Theo PT: \(n_{H_2}=\dfrac{1}{2}n_{C_2H_5OH}+\dfrac{1}{2}n_{CH_3COOH}=\dfrac{4,958}{24,79}=0,2\left(mol\right)\left(2\right)\)
Từ (1) và (2) \(\Rightarrow n_{C_2H_5OH}=n_{CH_3COOH}=0,2\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{C_2H_5OH}=\dfrac{0,2.46}{21,2}.100\%\approx43,4\%\\\%m_{CH_3COOH}\approx56,6\%\end{matrix}\right.\)
a) \(n_{H_2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)
PTHH: 2Na + 2H2O --> 2NaOH + H2
_____0,1<----------------0,1<------0,05
=> mNa = 0,1.23 = 2,3 (g)
=> \(\left\{{}\begin{matrix}\%Na=\dfrac{2,3}{4,7}.100\%=48,936\%\\\%Mg=100\%-48,936\%=51,064\%\end{matrix}\right.\)
b)
\(n_{Mg}=\dfrac{4,7-2,3}{24}=0,1\left(mol\right)\)
PTHH: Mg + 2HCl --> MgCl2 + H2
______0,1-->0,2-------------->0,1
2Na + 2HCl --> 2NaCl + H2
0,1-->0,1-------------->0,05
=> mHCl = (0,1+0,2).36,5 = 10,95 (g)
=> \(C\%\left(HCl\right)=\dfrac{10,95}{200}.100\%=5,475\%\)
=> VH2 = (0,1 + 0,05).22,4 = 3,36 (l)
a.
Số mol H2: nH2=2,24/22,4 =0,1 mol
Vì chỉ có Ca tác dụng với nước nên:
PT: Ca + 2H2O-->Ca(OH)2 + H2
0,1<--------------------------0,1 mol
Khối lượng Ca: mCa=0,1.40=4 g
mMg=8,8-4=4,8 g
b.
Trong hỗn hợp A:
nCa=4/40=0,1 mol
nMg=4,8/24=0,2 mol
PT:
Ca + 2HCl------->CaCl2 + H2
0,1-----------------------------0,1 mol
Mg + 2HCl-------->MgCl2 + H2
0,2--------------------------------0,2 mol
nH2=0,1 + 0,2=0,3 mol
Thể tích H2:
VH2=0,3.22,4=6,72 l
\(n_{H_2}=\dfrac{3,36}{22,4}=0,15mol\\ a)ZnO+H_2SO_4\rightarrow ZnSO_4+H_2O\\ Fe+H_2SO_4\rightarrow FeSO_4+H_2\)
0,15 0,15 0,15 0,15
\(b)m_{Fe}=0,15.56=8,4g\\ m_{ZnO}=16,5-8,4=8,1g\\ c)n_{ZnO}=\dfrac{8,1}{81}=0,1mol\\ ZnO+H_2SO_4\rightarrow ZnSO_4+H_2O\)
0,1 0,1 0,1 0,1
\(V_{ddH_2SO_4}=\dfrac{0,15+0,1}{2}=0,125M\\ d)Fe+CuSO_4\rightarrow FeSO_4+Cu\)
0,15 0,15 0,15 0,15
\(m_{rắn}=m_{ZnO}+m_{Cu}=8,1+0,15.64=17,7g\)
a) \(n\left(NaOH\right)=0,4.2=0,8\left(mol\right)\)
\(CH_3COOH+NaOH\rightarrow CH_3COONa+H_2O\left(1\right)\)
\(\left(1\right)\Rightarrow n\left(CH_3COOH\right)=0,8\left(mol\right)\)
\(m\left(CH_3COOH\right)=0,8.60=48\left(g\right)\)
\(m\left(C_2H_5OH\right)=57,2-48=9,2\left(g\right)\)
b) \(2CH_3COOH+2Na\rightarrow2CH_3COONa+H_2\uparrow\left(2\right)\)
\(2C_2H_5OH+2Na\rightarrow2C_2H_5ONa+H_2\uparrow\left(3\right)\)
\(n\left(C_2H_5OH\right)=\dfrac{9,2}{46}=0,2\left(mol\right)\)
\(\left(2\right)\Rightarrow n\left(H_2\right)=\dfrac{0,8}{2}=0,4\left(mol\right)\)
\(\left(3\right)\Rightarrow n\left(H_2\right)=\dfrac{0,2}{2}=0,1\left(mol\right)\)
Tổng số mol \(H_2=0,4+0,1=0,5\left(mol\right)\)
\(V\left(H_2\right)=0,5.24,79=12,4\left(lít\right)\)