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\(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\)
a) PTHH : \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)
Theo Pt : \(n_{Fe}=n_{H2SO4}=n_{FeSO4}=n_{H2}=0,2\left(mol\right)\)
b) \(V_{H2\left(dktc\right)}=0,2.22,4=4,48\left(l\right)\)
c) \(C_{MddH2SO4}=\dfrac{0,2}{0,2}=1\left(M\right)\)
d) \(m_{muối}=m_{FeSO4}=0,2.152=30,4\left(g\right)\)
a, PT: \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)
b, Ta có: \(n_{Fe}=\dfrac{19,6}{56}=0,35\left(mol\right)\)
Theo PT: \(n_{H_2}=n_{Fe}=0,35\left(mol\right)\Rightarrow V_{H_2}=0,35.22,4=7,84\left(l\right)\)
c, \(n_{H_2SO_4}=n_{Fe}=0,35\left(mol\right)\Rightarrow C_{M_{H_2SO_4}}=\dfrac{0,35}{0,2}=1,75\left(M\right)\)
d, \(n_{FeSO_4}=n_{Fe}=0,35\left(mol\right)\Rightarrow m_{FeSO_4}=0,35.152=53,2\left(g\right)\)
e, \(C_{M_{FeSO_4}}=\dfrac{0,35}{0,2}=1,75\left(M\right)\)
d, \(n_{H_2SO_4}=0,25.1,6=0,4\left(mol\right)\)
Xét tỉ lệ: \(\dfrac{n_{Fe}}{1}< \dfrac{n_{H_2SO_4}}{1}\), ta được H2SO4 dư.
Theo PT: \(n_{H_2SO_4\left(pư\right)}=n_{Fe}=0,35\left(mol\right)\)
\(\Rightarrow n_{H_2SO_4\left(dư\right)}=0,4-0,35=0,05\left(mol\right)\)
\(\Rightarrow m_{H_2SO_4\left(dư\right)}=0,05.98=4,9\left(g\right)\)
Bài 4 :
\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)
a) Pt : \(Zn+2HCl\rightarrow ZnCl_2+H_2|\)
1 2 1 1
0,2 0,4 0,2 0,2
b) \(n_{H2}=\dfrac{0,2.1}{1}=0,2\left(mol\right)\)
\(V_{H2\left(dktc\right)}=0,2.22,4=4,48\left(l\right)\)
c) \(n_{HCl}=\dfrac{0,2.2}{1}=0,4\left(mol\right)\)
⇒ \(m_{HCl}=0,4.36,5=14,6\left(g\right)\)
\(C_{ddHCl}=\dfrac{14,6.100}{100}=14,6\)0/0
d) \(n_{ZnCl2}=\dfrac{0,2.1}{1}=0,2\left(mol\right)\)
⇒ \(m_{ZnCl2}=0,2.136=27,2\left(g\right)\)
\(m_{ddspu}=13+100-\left(0,2.2\right)=112,6\left(g\right)\)
\(C_{ZnCl2}=\dfrac{27,2.100}{112,6}=24,16\)0/0
Chúc bạn học tốt
Bài 3 :
\(n_{Mg}=\dfrac{12}{24}=0,5\left(mol\right)\)
a) Pt : \(Mg+H_2SO_4\rightarrow MgSO_4+H_2|\)
1 1 1 1
0,5 0,5 0,5 0,5
b) \(n_{H2}=\dfrac{0,5.1}{1}=0,5\left(mol\right)\)
\(V_{H2\left(dktc\right)}=0,5.22,4=11,2\left(l\right)\)
c) \(n_{H2SO4}=\dfrac{0,5.1}{1}=0,5\left(mol\right)\)
⇒ \(m_{H2SO4}=0,5.98=49\left(g\right)\)
\(C_{ddH2SO4}=\dfrac{49.100}{200}=24,5\)0/0
d) \(n_{MgSO4}=\dfrac{0,5.1}{1}=0,5\left(mol\right)\)
⇒ \(m_{MgSO4}=0,5.120=60\left(g\right)\)
\(m_{ddspu}=12+200-\left(0,5.2\right)=211\left(g\right)\)
\(C_{MgSO4}=\dfrac{60.100}{211}=28,44\)0/0
Chúc bạn học tốt
Mg+ 2HCl→ MgCl2+ H2
(mol) 0,1 0,2 0,1 0,1
a) \(n_{Mg}=\dfrac{m}{M}=\dfrac{2,4}{24}=0,1\left(mol\right)\)
→\(V_{H_2}=n.22,4=0,1.22,4=2,24\left(lít\right)\)
b) Đổi: 100ml=0,1 lít
\(C_{M_{HCl}}=\dfrac{n}{V}=\dfrac{0,2}{0,1}=2M\)
c) \(m_{MgCl_2}=n.M=0,1.95=9,5\left(g\right)\)
đổi 500ml = 0,5l
n(CH\(_3\)COOH)\(_2\)Mg= \(\dfrac{14,2}{142}\)= 0,1mol
2CH3COOH + Mg \(\rightarrow\) (CH3COO)2Mg + H2
0,2mol 0,1mol 0,1mol
a/ CCH3COOH= \(\dfrac{0,2}{0,5}\)=0,4M
b/ VH\(_2\) 0,1 . 22,4 = 2,24l
c/ nCH\(_3\)COOH= 0,2mol
CH3COOH + NaOH \(\rightarrow\) CH3COONa + H2
0,2mol 0,2mol
V\(_{dd_{NaOH}}\)= \(\dfrac{0,2}{0,5}\)= 0,4l
\(n_{K_2SO_3}=\dfrac{15.8}{158}=0.1\left(mol\right)\)
\(K_2SO_3+H_2SO_4\rightarrow K_2SO_4+SO_2+H_2O\)
\(0.1...........................0.1..........0.1\)
\(V_{SO_2}=0.1\cdot22.4=2.24\left(l\right)\)
\(C_{M_{K_2SO_4}}=\dfrac{0.1}{0.2}=0.5\left(M\right)\)
a) \(n_{Al}=\dfrac{10,8}{27}=0,4\left(mol\right)\)
PTHH: 2Al + 3H2SO4 --> Al2(SO4)3 + 3H2
0,4-->0,6---------->0,2------->0,6
=> \(C_{M\left(dd.H_2SO_4\right)}=\dfrac{0,6}{0,15}=4M\)
b) VH2 = 0,6.22,4 = 13,44 (l)
c) \(C_{M\left(Al_2\left(SO_4\right)_3\right)}=\dfrac{0,2}{0,15}=\dfrac{4}{3}M\)
\(n_{Al}=\dfrac{10,8}{27}=0,4\left(mol\right)\\
pthh:2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
0,4 0,6 0,2 0,6
\(C_M_{H_2SO_4}=\dfrac{0,6}{0,15}=4M\\ V_{H_2}=0,622,4=13,44L\)
\(C_M=\dfrac{0,2}{0,15}=1,3M\)
nMg = 2,4/24 = 0,1 (mol)
a/ Mg + H2SO4 ------> MgSO4 + H2
b/ Từ PTHH ta suy ra nH2 = nMg = 0,1 (mol)
Suy ra \(V_{H_2}=22,4\times0,1=2,24\left(l\right)\)
c/ Từ PTHH suy ra nH2SO4 = nMg = 0,1 mol
Suy ra \(C_{M_{H2SO4}}=\frac{0,1}{\frac{200}{1000}}=0,5M\)
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