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\(a,n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)
PTHH: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
0,2-->0,4----------------->0,2
\(maxV_{H_2}=0,2.22,4=4,48\left(l\right)\\ b,C_{M\left(HCl\right)}=\dfrac{0,4}{0,2}=2M\\ c,n_{Cl}=m_{HCl}=0,4\left(mol\right)\\ BTKL:m_{muối}=m_{KL}+m_{Cl}=5,5+0,4.35,5=19,7\left(g\right)\)
a,\(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)
PTHH: Fe + H2SO4 → FeSO4 + H2
Mol: 0,1 0,1 0,1 0,1
b,\(V_{H_2}=0,1.22,4=2,24\left(l\right)\)
c,\(C_{M_{ddH_2SO_4}}=\dfrac{0,1}{0,2}=0,5M\)
d,\(C_{M_{ddFeSO_4}}=\dfrac{0,1}{0,2}=0,5M\)
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
=> mMgCl2 = 0,1.95 = 9,5 (g)
b) \(C_{M\left(HCl\right)}=\dfrac{0,2}{0,2}=1M\)
- PT: a, \(2KMnO_4+16HCl\rightarrow2KCl+2MnCl_2+5Cl_2+8H_2O\) (1)
\(MnO_2+4HCl_đ\underrightarrow{t^o}MnCl_2+Cl_2+2H_2O\) (2)
- Ta có: \(n_{HCl\left(1\right)}=n_{HCl\left(2\right)}=0,2.2=0,4\left(mol\right)\)
Theo PT (1): \(n_{Cl_2}=\dfrac{5}{16}n_{HCl\left(1\right)}=0,125\left(mol\right)\Rightarrow V_1=0,125.22,4=2,8\left(l\right)\)
(2): \(n_{Cl_2\left(2\right)}=\dfrac{1}{4}n_{HCl\left(2\right)}=0,1\left(mol\right)\Rightarrow V_2=0,1.22,4=2,24\left(l\right)\)
\(n_{Mg}=\dfrac{12}{24}=0,5\left(mol\right)\)
PT: \(Mg+H_2SO_4\rightarrow MgSO_4+H_2\)
a, \(n_{H_2}=n_{Mg}=0,5\left(mol\right)\)
\(\Rightarrow V_{H_2}=0,5.22,4=11,2\left(l\right)\)
b, \(n_{H_2SO_4}=n_{Mg}=0,5\left(mol\right)\)\(\Rightarrow C_{M_{H_2SO_4}}=\dfrac{0,5}{0,2}=2,5\left(M\right)\)
