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\(n_{HCl}=\dfrac{50.18,25}{100.36,5}=0,25(mol)\\ K_2CO_3+2HCl\to 2KCl+H_2O+CO_2\uparrow\\ a,n_{CO_2}=0,125(mol)\\ \Rightarrow V_{CO_2}=0,125.22,4=2,8(l)\\ b,n_{K_2CO_3}=0,125(mol)\\ \Rightarrow m_{K_2CO_3}=0,125.138=17,25(g)\\ c,n_{KCl}=0,25(mol)\\ \Rightarrow C\%_{KCl}=\dfrac{0,25.74,5}{17,25+50-0,125.44}.100\%=30,16\%\)
PTHH: \(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)
Ta có: \(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}n_{Fe}=n_{FeCl_2}=n_{H_2}=0,2\left(mol\right)\\n_{HCl}=0,4\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{Fe}=0,2\cdot56=11,2\left(g\right)\\m_{FeCl_2}=0,2\cdot127=25,4\left(g\right)\\m_{ddHCl}=\dfrac{0,4\cdot36,5}{10\%}=146\left(g\right)\\m_{H_2}=0,2\cdot2=0,4\left(g\right)\end{matrix}\right.\)
Mặt khác: \(m_{dd}=m_{Fe}+m_{ddHCl}-m_{H_2}=156,8\left(g\right)\) \(\Rightarrow C\%_{FeCl_2}=\dfrac{25,4}{156,8}\cdot100\%\approx16,2\%\)
a) PTHH: Zn + 2HCl -> ZnCl2 + H2
b) theo đề ra có : mZn = 6,5g
=> nZn = 0,1mol
pt:
Zn + 2HCl -> ZnCl2 + H2
1mol.....2mol.......1mol.......1mol
0,1mol..0,2mol.....0.1mol....0,1mol
theo pt: nZn = nH2 = 0,1mol
=> VH2 = 0,1.22,4 = 2,24l
vậy ....
c) theo pt 2nZn = nHCl = 0,2mol
=> CM = \(\dfrac{0,2}{0,1}=2M\)
vậy ...
\(n_{CO_2}=\dfrac{2,24}{22,4}=0,1mol\)
a)\(CaCO_3+2HCl\rightarrow CaCl_2+H_2O+CO_2\)
0,1 0,2 0,1 0,1 0,1
b)\(m_{HCl}=0,1\cdot36,5=3,65\left(g\right)\)
\(a\%=\dfrac{3,65}{100}\cdot100\%=3,65\%\)
c)\(m_{CaCO_3}=0,1\cdot100=10\left(G\right)\)
\(\Rightarrow\%m_{CaCO_3}=\dfrac{10}{16}\cdot100\%=62,5\%\)
\(\Rightarrow\%m_{CaCl_2}=100\%-62,5\%=37,5\%\)
d)\(m_{CaCl_2}=0,1\cdot111=11,1\left(g\right)\)
\(m_{H_2O}=0,1\cdot18=1,8\left(g\right)\)
\(m_{ddsau}=10+100-0,1\cdot44-1,8=103,8\left(g\right)\)
\(\Rightarrow C\%=\dfrac{11,1}{103,8}\cdot100\%=10,7\%\)
200ml = 0,2l
\(n_{Zn}=\dfrac{19,5}{65}=0,3\left(mol\right)\)
Pt : \(Zn+2HCl\rightarrow ZnCl_2+H_2|\)
1 2 1 1
0,3 0,6 0,3 0,3
a) \(n_{ZnCl2}=\dfrac{0,3.1}{1}=0,3\left(mol\right)\)
\(C_{M_{ZnCl2}}=\dfrac{0,3}{0,2}=1,5\left(M\right)\)
b) \(n_{H2}=\dfrac{0,3.1}{1}=0,3\left(mol\right)\)
\(V_{H2\left(dktc\right)}=0,3.22,4=6,72\left(l\right)\)
c) Pt : \(NaOH+HCl\rightarrow NaCl+H_2O|\)
1 1 1 1
0,6 0,6
\(n_{NaOH}=\dfrac{0,6.1}{1}=0,6\left(mol\right)\)
\(m_{NaOH}=0,6.40=24\left(g\right)\)
\(m_{ddNaOH}=\dfrac{24.100}{20}=120\left(g\right)\)
Chúc bạn học tốt
a) \(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)
\(m_{H_2SO_4}=147.10\%=14,7\left(g\right)\Rightarrow n_{H_2SO_4}=\dfrac{14,7}{98}=0,15\left(mol\right)\)
PTHH: Zn + H2SO4 → ZnSO4 + H2
Mol: 0,1 0,1 0,1 0,1
Ta có: \(\dfrac{0,1}{1}< \dfrac{0,15}{1}\) ⇒ Zn hết, H2SO4 dư
b) \(V_{H_2}=0,1.22,4=2,24\left(l\right)\)
c) mdd sau pứ = 6,5 + 147 - 0,1.2 = 153,3 (g)
\(C\%_{ddZnSO_4}=\dfrac{0,1.161.100\%}{153,3}=10,502\%\)
\(C\%_{ddH_2SO_4dư}=\dfrac{\left(0,15-0,1\right).98.100\%}{153,3}=3,196\%\)
\(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)
a. \(Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)
0,1----0,2----------------------0,1
b. \(V_{H_2}=0,1\cdot22,4=2,24l\)
c. \(CM_{HCl}=\dfrac{0,2}{0,25}=0,8\)
d. \(2HCl+Ca\left(OH\right)_2\rightarrow CaCl_2+2H_2O\)
---0,2----------0,1
\(V_{Ca\left(OH\right)_2}=\dfrac{0,1}{2}=0,05l=50ml\)
a) Zn + 2HCl -> ZnCl2 + H2
......0,1.......0,2..........0.1.........0,1(mol)
b) nZn = 6.5/65=0.1(mol)
VH2 = 0,1.22,4 = 2,24l
c)CM(HCl)=0.2/0.25=0.8(M)
d) Ca(OH)2 + 2 HCL --> CaCl2 + 2 H2O
..........0.1.................0.2...................................................(mol)
V=0.1/2=0.5(l)
a)Zn +2HCl------>ZnCl2 +H2
\(b.\)Ta có
n\(_{Zn}=\frac{6.6}{65}=0,1mol\)
Theo Pthh
n\(_{HCl}=2n_{Zn}=0,2mol\)
C\(_M\left(HCl\right)=\frac{0,1}{0,1}=0,5\left(M\right)\)
c) Tính nồng độ mol của dd sau pư nha bạn...Bài này ko đủ dữ kiện để tính nồng độ phần tram
Theo pthh
n\(_{Z_{ }nCl2}=n_{Zn}=0,1mol\)
C\(_{M\left(ZnCl2\right)}=\frac{0,1}{0,1}=1\left(M\right)\)
d) Theo pthh
n\(_{H2}=n_{Zn}=0,1mol\)
V\(_{H2\left(đktc\right)}=0,1.22,4=2,24\left(l\right)\)
Chúc bạn học tốt
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