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a, \(Fe+2HCl\rightarrow FeCl_2+H_2\)
Ta có: \(n_{Fe}=\dfrac{14}{56}=0,25\left(mol\right)\)
Theo PT: \(\left\{{}\begin{matrix}n_{HCl}=2n_{Fe}=0,5\left(mol\right)\\n_{H_2}=n_{Fe}=0,25\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow V_{H_2}=0,25.24,79=6,1975\left(l\right)\)
\(a=C_{M_{HCl}}=\dfrac{0,5}{0,1}=5\left(M\right)\)
b, Theo PT: \(n_{FeCl_2}=n_{Fe}=0,25\left(mol\right)\)
Ta có: \(n_{AgNO_3}=0,4.1,3=0,52\left(mol\right)\)
PT: \(2AgNO_3+FeCl_2\rightarrow Fe\left(NO_3\right)_2+2AgCl_{\downarrow}\)
______0,5______0,25______0,25________0,5 (mol)
\(AgNO_3+Fe\left(NO_3\right)_2\rightarrow Fe\left(NO_3\right)_3+Ag_{\downarrow}\)
0,02______0,02________0,02________0,02 (mol)
⇒ m = mAgCl + mAg = 0,5.143,5 + 0,02.108 = 73,91 (g)
- Dd sau pư gồm: Fe(NO3)3: 0,02 (mol) và Fe(NO3)2: 0,25 - 0,02 = 0,23 (mol)
\(\Rightarrow\left\{{}\begin{matrix}C_{M_{Fe\left(NO_3\right)_3}}=\dfrac{0,02}{0,1+0,4}=0,04\left(M\right)\\C_{M_{Fe\left(NO_3\right)_2}}=\dfrac{0,23}{0,1+0,4}=0,46\left(M\right)\end{matrix}\right.\)
\(Fe+2HCl->FeCl_2+H_2\\ a.V=\dfrac{14}{56}\cdot22,4=5,6\left(L\right)\\ a=\dfrac{\dfrac{14}{56}\cdot2}{0,1}=5\left(M\right)\\ b.n_{AgNO_3}=0,4\cdot1,3=0,52mol\\ FeCl_2+AgNO_3->Fe\left(NO_3\right)_2+AgCl\\ Fe\left(NO_3\right)_2+AgNO_3->Ag+Fe\left(NO_3\right)_3\\ m=0,25\cdot143,5+0,25\cdot108=62,875\left(g\right)\\ C_{M\left(AgNO_3\right)}=\dfrac{0,02}{0,5}=0,04M\\ C_{M\left(Fe\left(NO_3\right)_3\right)}=\dfrac{0,25}{0,5}=0,5M\)
Câu 1 :
\(n_{H_2SO_4}=0.2\cdot0.1=0.02\left(mol\right)\)
\(n_{KOH}=0.3\cdot0.1=0.03\left(mol\right)\)
\(2KOH+H_2SO_4\rightarrow K_2SO_4+2H_2O\)
\(2..............1\)
\(0.03............0.02\)
Lập tỉ lệ : \(\dfrac{0.03}{2}< \dfrac{0.02}{1}\) \(\Rightarrow H_2SO_4dư\)
\(n_{K_2SO_4}=\dfrac{0.03}{2}=0.015\left(mol\right)\)
\(n_{H_2SO_{4\left(dư\right)}}=0.02-0.015=0.005\left(mol\right)\)
\(V_{ddX}=0.2+0.3=0.5\left(l\right)\)
\(C_{M_{K_2SO_4}}=\dfrac{0.015}{0.5}=0.03\left(M\right)\)
\(C_{M_{H_2SO_4\left(dư\right)}}=\dfrac{0.005}{0.5}=0.01\left(M\right)\)
Câu 2 :
\(n_{H_2}=\dfrac{0.672}{22.4}=0.03\left(mol\right)\)
Hai kim loại ở 2 chu kỳ liên kết thuộc nhóm IA => Đặt CT chung là : M
\(M+H_2O\rightarrow MOH+\dfrac{1}{2}H_2\)
\(0.06............................0.03\)
\(M_M=\dfrac{0.6}{0.06}=10\)\(\Rightarrow9< 10< 23\)
Hai kim loại là : Li và Na
\(n_{Li}=a\left(mol\right),n_{Na}=b\left(mol\right)\)
\(m_{hh}=9a+23b=0.6\left(g\right)\left(1\right)\)
\(n_{H_2}=0.5a+0.5b=0.03\left(mol\right)\left(2\right)\)
\(\left(1\right),\left(2\right):a=0.056,b=0.004\)
\(\%m_{Li}=\dfrac{0.056\cdot9}{0.6}\cdot100\%=84\%\)
\(\%m_{Na}=100-84=16\%\)
a) \(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
PTHH: R + 2HCl --> RCl2 + H2
____0,15<-----------------0,15
=> \(M_R=\dfrac{3,6}{0,15}=24\left(Mg\right)\)
b)
PTHH: Mg + 2HCl --> MgCl2 + H2
__________0,3<-----0,15<---0,15
=> \(V=\dfrac{0,3}{2}=0,15\left(l\right)=150\left(ml\right)\)
\(C_{M\left(MgCl_2\right)}=\dfrac{0,15}{0,15}=1M\)
