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\(\text{2K + 2H2O }\rightarrow\text{2KOH + H2}\)
0,2__________0,2________0,1
\(\text{KOH + HCl}\rightarrow\text{KCl + H2O}\)
0,2_____0,2___________0,2
nKCl= \(\frac{14,8}{74,5}\)= 0,2 (mol)
\(\rightarrow\)mK= 0,2 .39= 7,8 (g)
\(\rightarrow\)VH2= 0,1.22,4= 2,24 (l)
VHCl=\(\frac{0,2}{0,5}\)=0,4M
Gọi : \(\left\{{}\begin{matrix}n_{Al_2O_3}=a\left(mol\right)\\n_{Zn}=b\left(mol\right)\end{matrix}\right.\)⇒ 102a + 65b = 2,505(1)
\(Al_2O_3 + 6HCl \to 2AlCl_3 + 3H_2O\\ Zn + 2HCl \to ZnCl_2 + H_2\)
Muối gồm : \(\left\{{}\begin{matrix}AlCl_3:2a\left(mol\right)\\ZnCl_2:b\left(mol\right)\end{matrix}\right.\)⇒ 133,5.2a + 136b = 6,045(2)
Từ (1)(2) suy ra : a = 0,015 ; b = 0,015
Vậy :
\(\%m_{Al_2O_3} = \dfrac{0,015.102}{2,505}.100\% = 61,08\%\\ \%m_{Zn} = 100\% - 61,08\% = 38,92\%\)
Theo PTHH : \(n_{HCl} = 6a + 2b = 0,12(mol)\\ \Rightarrow C\%_{HCl} = \dfrac{0,12.36,5}{200}.100\% = 2,19\%\)
\(n_{H_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\\ PTHH:2Al+6HCl\rightarrow2AlCl_3+3H_2\\ Fe+2HCl\rightarrow FeCl_2+3H_2\\ Đặt:n_{Al}=a\left(mol\right);n_{Fe}=b\left(mol\right)\left(a,b>0\right)\\ \Rightarrow\left\{{}\begin{matrix}27a+56b=11\\1,5a+b=0,4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,2\\b=0,1\end{matrix}\right.\\ a,n_{HCl}=2.n_{H_2}=2.0,4=0,8\left(mol\right)\\ \Rightarrow V_{ddHCl}=\dfrac{0,8}{8}=0,1\left(l\right)\\ b,FeCl_2+2AgNO_3\rightarrow Fe\left(NO_3\right)_2+2AgCl\downarrow\\ AlCl_3+3AgNO_3\rightarrow Al\left(NO_3\right)_3+3AgCl\downarrow\\ n_{AgCl}=n_{AgNO_3}=3.n_{AlCl_3}+2.n_{FeCl_2}=3.a+2.b=3.0,2+2.0,1=0,8\left(mol\right)\\ \Rightarrow a=\dfrac{170.0,8}{250}.100=54,4\%\\ b=m_{\downarrow}=m_{AgCl}=0,8.143,5=114,8\left(g\right)\)
1)2Al+6HCl ->2Al2Cl3+3H2
Fe+2HCl->FeCl2+H2
Gọi số mol của Al là x;Fe là y
ta có 2x*23+56y=8.3
3x+y=5.6/22.4
giải ra là xong hết bài 1 r nha
\(n_{Zn}=a\left(mol\right),n_{Al}=b\left(mol\right)\)
\(n_{H_2}=\dfrac{8.96}{22.4}=0.4\left(mol\right)\)
\(Zn+2HCl\rightarrow ZnCl_2+H_2\)
\(2Al+6HCl\rightarrow2AlCl_3+H_2\)
\(n_{H_2}=a+1.5b=0.4\left(mol\right)\left(1\right)\)
\(m_{Muối}=m_{ZnCl_2}+m_{AlCl_3}=136a+133.5b=40.3\left(g\right)\left(2\right)\)
\(\left(1\right),\left(2\right):a=0.1,b=0.2\)
\(m_{hh}=0.1\cdot65+0.2\cdot27=11.9\left(g\right)\)
\(\%Zn=\dfrac{0.1\cdot65}{11.9}\cdot100\%=54.62\%\)
\(\%Al=100-54.62=45.38\%\)
a)
Ca + 2H2O → Ca(OH)2 + H2↑ (1)
Ca(OH)2 + 2HCl → CaCl2 + 2H2O (2)
nHCl = 0,5 . 0,2 = 0,1 mol
Theo (2): nCa(OH)2 =\(\frac{1}{2}\)nHCl = 0,05 mol
Theo (1): nCa = nCa(OH)2 = 0,05 mol
mCa = 0,05 . 40 = 2 (g)
b)
Theo (1): nH2 = nCa = 0,05 mol
mdd sau p.ứ (1) = mCa + mH2O - mH2
= 2 + 200 - 0,05 . 2
= 201,9 (g)
\(C\%_{Ca\left(OH\right)2}=\frac{0,05.74}{201,9}.100\%=1,83\%\)
\(2NaBr+Cl_2\rightarrow2NaCl+Br_2\left(1\right)\\ m_{giảm}=m_{Br_2}-m_{Cl_2}\\ \Leftrightarrow n_{NaCl\left(1\right)}=n_{NaBr\left(1\right)}=\dfrac{13,35}{160-71}=0,15\left(mol\right)\\ \Rightarrow\%m_{NaBr}=\dfrac{103.0,15}{42,6}.100\approx36,268\%\\ \Rightarrow\%m_{NaCl}\approx63,732\%\)
Bổ sung:
\(C\%_{ddNaBr\left(trongA\right)}=\dfrac{0,15.103}{200}.100=7,725\%\\ C\%_{ddNaCl\left(trongA\right)}=\dfrac{42,6-0,15.103}{200}.100=13,575\%\)
2Na + 2H2O \(\rightarrow\) 2NaOH + H2
0,4___________0,4______0,2
NaOH + HCl\(\rightarrow\) NaCl + H2O
0,4_____0,4____0,4
nNaCl= 0,4 (mol)
\(\rightarrow\) mNa= 9,2 (g)
mdd sau phản ứng= 9,2 + 100 - (0,2 . 2)=108,8 (g)
mNaOH= 16(g)
\(\rightarrow\) C%NaOH = \(\frac{16}{108,8}\).100= 14,7%
mddHCl=\(\frac{\text{0,4.36,5.100}}{10}\)=146 (g)