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\(a,Zn+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Zn+H_2\\ n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\\ b,n_{CH_3COOH}=2.0,2=0,4\left(mol\right)\\ C_{MddCH_3COOH}=\dfrac{0,4}{0,4}=1\left(M\right)\\ c,CH_3COOH+KOH\rightarrow CH_3COOK+H_2O\\ n_{CH_3COOK}=n_{CH_3COOH}=0,4\left(mol\right)\\ V_{ddCH_3COOK}=400+400=800\left(ml\right)=0,8\left(l\right)\\ C_{MddCH_3COOK}=\dfrac{0,4}{0,8}=0,5\left(M\right)\)
\(Zn+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Zn+H_2\uparrow\)
0,4 0,2
\(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
\(b,C_{M_{CH_3COOH}}=\dfrac{0,4}{0,4}=1M\)
\(c,CH_3COOH+KOH\rightarrow CH_3COOK+H_2O\)
0,4 0,4 0,4
\(C_{M_{CH_3COOK}}=\dfrac{0,4}{0,4}=1M\)
a, PT: \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)
Ta có: \(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
Theo PT: \(n_{Zn}=n_{H_2SO_4}=n_{ZnCl_2}=n_{H_2}=0,15\left(mol\right)\)
b, mZn = 0,15.65 = 9,75 (g)
c, CM (H2SO4) = 0,15/0,05 = 3 M
d, mZnSO4 = 0,15.161 = 24,15 (g)
Bạn tham khảo nhé!
a) PTHH: Zn + H2SO4 -> ZnSO4 + H2
nH2= 0,15(mol)
=> nZn=nH2SO4=nZnSO4=nH2=0,15(mol)
b) mZn=0,15.65=9,75(g)
c) CMddH2SO4= 0,15/ 0,05=3(M)
d) mZnSO4= 161. 0,15=24,15(g)
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: \(n_{H_2SO_4\left(pư\right)}=n_{Fe}=0,35\left(mol\right)\)
\(\Rightarrow n_{H_2SO_4\left(dư\right)}=0,4-0,35=0,05\left(mol\right)\)
\(\Rightarrow m_{H_2SO_4\left(dư\right)}=0,05.98=4,9\left(g\right)\)
a) \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
b) \(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)
PTHH: Zn + 2HCl --> ZnCl2 + H2
0,1-->0,2------------->0,1
=> \(C_{M\left(HCl\right)}=\dfrac{0,2}{0,2}=1M\)
c)
PTHH: 2H2 + O2 --to--> 2H2O
0,1-->0,05
=> \(m_{O_2}=0,05.32=1,6\left(g\right)\)
a, \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
b, \(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
Theo PT: \(n_{Zn}=n_{H_2}=0,3\left(mol\right)\)
\(\Rightarrow m_{Zn}=0,3.65=19,5\left(g\right)\)
c, \(n_{HCl}=2n_{H_2}=0,6\left(mol\right)\)
\(\Rightarrow C_{M_{HCl}}=\dfrac{0,6}{0,1}=6\left(M\right)\)
a, \(Zn+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Zn+H_2\)
Bạn bổ sung thêm số liệu của khí thoát ra nhé.