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a, \(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
Theo PT: \(n_{Zn}=n_{H_2}=0,1\left(mol\right)\Rightarrow m_{Zn}=0,1.65=6,5\left(g\right)\)
b, \(n_{HCl}=2n_{H_2}=0,2\left(mol\right)\Rightarrow m_{ddHCl}=\dfrac{0,2.36,5}{3,65\%}=200\left(g\right)\)
\(n_{Zn}=\dfrac{13}{65}=0.2\left(mol\right)\)
\(Zn+2HCl\rightarrow ZnCl_2+H_2\)
\(0.2....................0.2..........0.2\)
\(m_{ZnCl_2}=0.2\cdot136=27.2\left(g\right)\)
\(V_{H_2}=0.2\cdot22.4=4.48\left(l\right)\)
a, Zn + 2HCl ---> ZnCl2 + H2
b, nZn=\(\dfrac{13}{65}=0,2mol\)
Ta có: 1 mol Zn ---> 1 mol H2
nên 0,2 mol Zn ---> 0,2 mol H2
VH2=0,2.22,4=4,48 mol
\(n_{H_2}=\dfrac{4.48}{22.4}=0.2\left(mol\right)\)
\(Zn+2HCl\rightarrow ZnCl_2+H_2\)
\(0.2.......0.4.......................0.2\)
\(m_{Zn}=0.2\cdot65=13\left(g\right)\)
\(C\%_{HCl}=\dfrac{0.4\cdot36.5}{200}\cdot100\%=7.3\%\)
\(n_{CuO}=\dfrac{24}{80}=0.3\left(mol\right)\)
\(CuO+H_2\underrightarrow{^{t^0}}Cu+H_2O\)
\(1..........1\)
\(0.3.........0.2\)
\(LTL:\dfrac{0.3}{1}>\dfrac{0.2}{1}\Rightarrow CuOdư\)
\(m_{CuO\left(dư\right)}=\left(0.3-0.2\right)\cdot64=6.4\left(g\right)\)
a) \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
nZn = \(\dfrac{13}{65}=0,2mol\)
Theo pt: nH2 = nZn = 0,2 mol
=> VH2 = 0,2.22,4 = 4,48 lít
b) Theo pt: nZnCl2 = nZn = 0,2 mol
=> mZnCl2 = 0,2.136 = 27,2g
\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right);n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\\a, Zn+2HCl\rightarrow ZnCl_2+H_2\\ b,V\text{ì}:\dfrac{0,2}{1}>\dfrac{0,1}{1}\Rightarrow Zn\text{dư}\\ \Rightarrow n_{Zn\left(p.\text{ứ}\right)}=n_{ZnCl_2}=n_{H_2}=0,1\left(mol\right)\\b, m_{Zn\left(p.\text{ứ}\right)}=0,1.65=6,5\left(g\right)\\ n_{HCl}=0,1.2=0,2\left(mol\right)\\ m_{HCl}=0,2.36,5=7,3\left(g\right)\\ d,m_{ZnCl_2}=136.0,1=13,6\left(g\right)\)
Fe+2HCl->FeCl2+H2
0,1---------------------0,1
2H2+O2-to>2H2O
0,1----0,05 mol
0,1--0,1
n Fe=\(\dfrac{5,6}{56}\)=0,1 mol
=>VH2=0,1.22,4=2,24l
=>mkk=0,05.29=1,45l
Theo gt ta có: $n_{Zn}=0,1(mol)$
a, $Zn+2HCl\rightarrow ZnCl_2+H_2$
b, Ta có: $n_{H_2}=0,1(mol)\Rightarrow V_{H_2}=2.24(l)$
c, Ta có: $n_{HCl}=2.n_{Zn}=0,2(mol)\Rightarrow m_{HCl}=7,3(g)$
n\(_{H_2}\) = \(\dfrac{2,24}{22,4}\)= 0,1 (mol)
a)
PTHH: Zn + 2HCl ----> ZnCl2 + H2\(\uparrow\)
mol: __0,1<-0,2<------------------0,1
b) m\(_{Zn}\)= 0,1 . 65 = 6,5 (g)
c)
PTHH: NaOH + HCl ----> NaCl + H2O
mol:__0,2<------0,2
m\(_{NaOH}\)= 0,2 . 40 = 8 (g)
mdd\(_{NaOH}\) = \(\dfrac{8.100}{15}\)= 53,33 (g)