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\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\\Zn+2HCl\rightarrow ZnCl_2+H_2\\ n_{HCl}=0,2.2=0,4\left(mol\right)\\ m_{ddHCl}=\dfrac{0,4.36,5.100}{10}=146\left(g\right)\\ n_{H_2}=n_{ZnCl_2}=n_{Zn}=0,2\left(mol\right)\\ V_{H_2\left(đktc\right)}=0,2.22,4=4,48\left(l\right)\\ C\%_{ddZnCl_2}=\dfrac{136.0,2}{13+146-0,2.2}.100\approx17,15\%\)
\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)
PTHH: Zn + 2HCl → ZnCl2 + H2
Mol: 0,2 0,4 0,2 0,2
\(V_{H_2}=0,2.22,4=4,48\left(l\right)\)
\(m_{ddHCl}=\dfrac{0,4.36,5.100}{10}=146\left(g\right)\)
mdd sau pứ = 13+146-0,2.2 = 158,6 (g)
\(C\%_{ddZnCl_2}=\dfrac{0,2.136.100\%}{158,6}=17,15\%\)
PTHH: \(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)
Ta có: \(n_{HCl}=\dfrac{200\cdot71\%}{36,5}=\dfrac{284}{73}\left(mol\right)\)
\(\Rightarrow n_{Fe}=n_{FeCl_2}=n_{H_2}=\dfrac{142}{73}\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}m_{Fe}=\dfrac{142}{73}\cdot56\approx108,93\left(g\right)\\m_{FeCl_2}=\dfrac{142}{73}\cdot127\approx247,04\left(g\right)\\m_{H_2}=\dfrac{142}{73}\cdot2\approx3,89\left(g\right)\\V_{H_2}=\dfrac{142}{73}\cdot22,4\approx43,57\left(l\right)\end{matrix}\right.\)
Mặt khác: \(m_{dd}=m_{Fe}+m_{ddHCl}-m_{H_2}=305,04\left(g\right)\)
\(\Rightarrow C\%_{FeCl_2}=\dfrac{247,04}{305,04}\cdot100\%\approx80,99\%\)
Fe + 2HCl ➝ FeCl2 + H2
mHCl = 200.71% = 142 (g) => nHCl = \(\dfrac{284}{73}\) (mol)
nFe = \(\dfrac{1}{2}\) nHCl = \(\dfrac{142}{73}\) (mol) => m ≃ 108,9 (g)
nH2 = nFe => V ≃ 43,57 (l)
nFeCl2 = nFe => C% ≃ 80%
(Mk nghĩ bạn nên kiểm tra lại đề vì số liệu không được đẹp cho lắm)
\(Zn+2HCl\rightarrow ZnCl_2+H_2\)
Theo PT: \(n_{Zn}=n_{H_2}=n_{ZnCl_2}=0,2\left(mol\right)\)
=> \(V_{H_2}=0,2.22,4=4,48\left(l\right)\)
\(m_{ZnCl_2}=0,2.136=27,2\left(g\right)\)
Ta có \(n_{Zn}=\dfrac{13}{65}=0.2\left(mol\right)\)
PTHH: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
0,2---->0,4----->0,2------->0,2(mol)
\(V_{H_2\left(đktc\right)}=0,2\cdot22,4=4,48\left(l\right)\)
\(m_{ZnCl_2}=0,2\cdot\left(65+35,5\cdot2\right)=27,2\left(g\right)\)
Tick nha
a)
$Fe + 2HCl \to FeCl_2 + H_2$
Theo PTHH :
$n_{Fe} = n_{H_2} = \dfrac{11,2}{22,4} = 0,5(mol)$
$A = 0,5.56 = 28(gam)$
b) $n_{HCl} = 2n_{H_2} = 1(mol)$
$m_{HCl} = 1.36,5 = 36,5(gam)$
c) $m_{dd\ HCl} = 36,5 : 20\% = 182,5(gam)$
$m_{dd\ sau\ pư} = 28 + 182,5 - 0,5.2 = 209,5(gam)$
$C\%_{FeCl_2} = \dfrac{0,5.127}{209,5}.100\% = 30,3\%$
a, \(Na_2CO_3+2HCl\rightarrow2NaCl+CO_2+H_2O\)
b,\(n_{Na_2CO_3}=\dfrac{21,2}{106}=0,2\left(mol\right)\)
Theo PT: \(n_{HCl}=2n_{Na_2CO_3}=0,4\left(mol\right)\)
\(\Rightarrow m_{HCl}=0,4.36,5=14,6\left(g\right)\)
c, \(n_{CO_2}=n_{Na_2CO_3}=0,2\left(mol\right)\)
\(\Rightarrow V_{CO_2}=0,2.22,4=4,48\left(l\right)\)
\(n_{H_2}=\dfrac{2.24}{22.4}=0.1\left(mol\right)\)
\(Zn+2HCl\rightarrow ZnCl_2+H_2\)
\(0.1.......0.2......................0.1\)
Chất rắn X : Cu
\(m_{Zn}=0.1\cdot65=6.5\left(g\right)\Rightarrow m_{Cu}=19.3-6.5=12.8\left(g\right)\)
\(n_{Cu}=\dfrac{12.8}{64}=0.2\left(mol\right)\)
\(C_{M_{HCl}}=\dfrac{0.2}{0.2}=1\left(M\right)\)
\(2Cu+O_2\underrightarrow{^{^{t^o}}}2CuO\)
\(0.2........0.1\)
\(m_{tăng}=m_{O_2}=0.1\cdot32=3.2\left(g\right)\)
a)
\(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
PTHH: Mg + 2HCl --> MgCl2 + H2
_____0,1<---0,2<-------0,1<---0,1
=> mHCl = 0,2.36,5 = 7,3 (g)
=> \(m_{ddHCl}=\dfrac{7,3.100}{7,3}=100\left(g\right)\)
mdd sau pư = 0,1.24 + 100 - 0,1.2 = 102,2 (g)
\(C\%\left(MgCl_2\right)=\dfrac{0,1.95}{102,2}.100\%=9,2955\%\)
b)
CTHH: AaOb
PTHH: \(A_aO_b+2bHCl->aACl_{\dfrac{2b}{a}}+bH_2O\)
____________0,2------->\(\dfrac{0,1a}{b}\)
=> \(\dfrac{0,1a}{b}\left(M_A+35,5.\dfrac{2b}{a}\right)=13,5\)
=> \(M_A=\dfrac{64b}{a}=\dfrac{2b}{a}.32\)
Nếu \(\dfrac{2b}{a}=1\) => MA = 32 (L)
Nếu \(\dfrac{2b}{a}=2\) => MA = 64(Cu)
a) \(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\)
PTHH: Fe + 2HCl → FeCl2 + H2
Mol: 0,2 0,4 0,2 0,2
b) \(V_{H_2}=0,2.22,4=4,48\left(l\right)\)
c) \(C\%_{ddHCl}=\dfrac{0,4.36,5.100\%}{300}=4,87\%\)
d) mdd sau pứ = 11,2 + 300 - 0,2.2 = 310,8 (g)
\(C\%_{ddFeCl_2}=\dfrac{0,2.127.100\%}{310,5}=8,17\%\)
\(a.n_{Zn}=\dfrac{13}{65}=0,2mol\\ Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)
0,2 0,4 0,2 0,2
\(V_{H_2}=0,2.24,79=4,958l\\ b.m_{ZnCl_2}=0,2.136=27,2g\\ c.m_{dd\left(sau.pư\right)}=\dfrac{0,4.36,5}{7,3}\cdot100+13-0,2.2=212,6g\\ C_{\%ZnCl_2}=\dfrac{27,2}{212,6}\cdot100=12,79\%\)
Em cảm ơn ạaaaaa