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\(n_{Al}=\dfrac{27}{27}=1\left(mol\right)\\ n_{HCl}=0,2.0,6=0,12\left(mol\right)\\ 2Al+6HCl\rightarrow2AlCl_3+3H_2\\ Vì:\dfrac{1}{2}>\dfrac{0,12}{6}\Rightarrow Aldư,tính.theo.n_{HCl}\\ \Rightarrow ddA:AlCl_3\\ n_{AlCl_3}=\dfrac{2}{6}.n_{HCl}=\dfrac{2}{6}.0,12=0,04\left(mol\right)\\ V_{ddA}=V_{ddHCl}=0,2\left(l\right)\\ C_{MddAlCl_3}=\dfrac{0,04}{0,2}=0,2\left(M\right)\\ n_{H_2}=\dfrac{3}{6}.n_{HCl}=\dfrac{3}{6}.0,12=0,06\left(mol\right)\\ V=V_{H_2\left(đktc\right)}=0,06.22,4=1,344\left(l\right)\)
\(n_{NaOH}=0,5.0,2=0,1\left(mol\right)\\ n_{H_2SO_4}=0,3.0,1=0,03\left(mol\right)\\ PTHH:2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\\ Vì:\dfrac{0,1}{2}>\dfrac{0,03}{1}\Rightarrow NaOHdư\\ \Rightarrow ddA:NaOH\left(dư\right),Na_2SO_4\\ n_{Na_2SO_4}=n_{H_2SO_4}=0,03\left(mol\right)\\ n_{NaOH\left(dư\right)}=0,1-0,03.2=0,04\left(mol\right)\\ V_{ddA}=V_{ddNaOH}+V_{ddH_2SO_4}=0,5+0,3=0,8\left(l\right)\\ C_{MddNaOH\left(dư\right)}=\dfrac{0,04}{0,8}=0,05\left(M\right)\\ C_{MddNa_2SO_4}=\dfrac{0,03}{0,8}=0,0375\left(M\right)\)
\(Gọi.V.là.thể.tích.dung.dịch.NaOH.0,2M\left(V>0\right)\left(lít\right)\\ \Rightarrow n_{NaOH\left(tổng\right)}=0,2V+0,5\left(mol\right)\\ V_{ddNaOH\left(tồng\right)}=V+0,5\left(l\right)\\ C_{MddNaOH\left(tổng\right)}=0,4\left(M\right)\\ \Leftrightarrow\dfrac{0,2V+0,5}{V+0,5}=0,4\\ \Leftrightarrow V=1,5\left(l\right)\)
\(S=2P+N=40\left(1\right)\\ 2P-N=12\left(2\right)\\ \left(1\right)+\left(2\right)\Rightarrow4P=52\Leftrightarrow P=13=E\\ \Rightarrow N=14\)
Bài 4:
a) \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
b) nHCl = 0,5.1 = 0,5 (mol)
\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)
Xét tỉ lệ: \(\dfrac{0,2}{1}< \dfrac{0,5}{2}\) => Zn hết, HCl dư
c)
PTHH: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
0,2---->0,4------>0,2
\(\left\{{}\begin{matrix}C_{M\left(HCl.dư\right)}=\dfrac{0,5-0,4}{0,5}=0,2M\\C_{M\left(ZnCl_2\right)}=\dfrac{0,2}{0,5}=0,4M\end{matrix}\right.\)
Bài 5:
a) \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)
b) \(n_{Fe}=\dfrac{22,4}{56}=0,4\left(mol\right)\); \(n_{H_2SO_4}=\dfrac{100.49\%}{98}=0,5\left(mol\right)\)
Xét tỉ lệ: \(\dfrac{0,4}{1}< \dfrac{0,5}{1}\) => Fe hết, H2SO4 dư
c)
PTHH: \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)
0,4---->0,4------->0,4--->0,4
\(\left\{{}\begin{matrix}m_{H_2SO_4\left(dư\right)}=\left(0,5-0,4\right).98=9,8\left(g\right)\\m_{FeSO_4}=0,4.152=60,8\left(g\right)\end{matrix}\right.\)
mdd sau pư = 22,4 + 100 - 0,4.2 = 121,6 (g)
\(\left\{{}\begin{matrix}C\%_{H_2SO_4\left(dư\right)}=\dfrac{9,8}{121,6}.100\%=8,06\%\\C\%_{FeSO_4}=\dfrac{60,8}{121,6}.100\%=50\%\end{matrix}\right.\)
Bài 6:
\(n_{HCl}=1.0,6=0,6\left(mol\right)\)
PTHH: \(A_2O_3+6HCl\rightarrow2ACl_3+3H_2O\)
0,1<----0,6
=> \(M_{A_2O_3}=\dfrac{10,2}{0,1}=102\left(g/mol\right)\Rightarrow M_A=27\left(g/mol\right)\)
=> A là Al
CTHH: Al2O3
\(n_{CuO\left(pư\right)}=\dfrac{48.80\%}{80}=0,48\left(mol\right)\)
\(n_{CuO\left(bđ\right)}=\dfrac{48}{80}=0,6\left(mol\right)\)
PTHH: \(CuO+H_2\underrightarrow{t^o}Cu+H_2O\)
0,48->0,48->0,48
Chất rắn sau pư gồm \(\left\{{}\begin{matrix}CuO:0,12\left(mol\right)\\Cu:0,48\left(mol\right)\end{matrix}\right.\)
=> mchất rắn sau pư = 0,12.80 + 0,48.64 = 40,32 (g)
\(V_{H_2}=0,48.22,4=10,752\left(l\right)\)
\(n_{CuO}=\dfrac{48}{80}=0,6\left(mol\right)\)
Vì H = 80%
=> \(\left\{{}\begin{matrix}n_{CuO\left(pư\right)}=0,6.80\%=0,48\left(mol\right)\\n_{CuO\left(chưa.pư\right)}=0,6-0,48=0,12\left(mol\right)\end{matrix}\right.\)
PTHH: CuO + H2 --to--> Cu + H2O
0,48--->0,48--->0,48
=> mchất rắn sau pư = 0,48.64 + 0,12.80 = 40,32 (g)
VH2 = 0,48.22,4 = 10,752 (l)
\(C7\\ CTTQ:R_aO_b\\ m_R=M_R.a\left(g\right)\\ m_O=16b\left(g\right)\\ HC:\dfrac{m_R}{m_O}=\dfrac{3}{8}\\ \Leftrightarrow\dfrac{M_R.a}{16b}=\dfrac{3}{8}\\ \Leftrightarrow\dfrac{a}{b}=\dfrac{48}{8M_R}=\dfrac{6}{M_R}\)
Giả sử:
+ Nếu R là Ca => a/b=6/40=3/20 (loại)
+ Nếu R là S => a/b= 6/32= 3/16 (loại)
+ Nếu R là C => a/b= 6/12=1/2 (a=1;b=2) => HC: CO2 (Nhận)
Vậy R là Cacbon (C)
\(C6\\ PTK_{hc}=64\left(đ.v.C\right)\\ CTTQ:S_xO_y\left(x,y:nguyên,dương\right)\\ \%m_S=100\%-50\%=50\%\\ x=\dfrac{64.50\%}{32}=1;y=\dfrac{64.50\%}{16}=2\\ Với:x=1;y=2\Rightarrow HC:SO_2\\ HC.có.1.nguyên.tử.lưu.huỳnh.và.2.nguyên.tử.oxi\\ Tên.gọi:Lưu.huỳnh.dioxit\left(khí.sunfuro\right)\\ Sinh.ra.khí.cho.kim.loại.phản.ứng.với.dung.dịch.H_2SO_{4\left(đ,nóng\right)}\\ Hoặc.sinh.ra.khi.đốt.lưu.huỳnh\)