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![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)
\(n_{HCl}=0,2.1=0,2\left(mol\right)\)
PTHH: Zn + 2HCl --> ZnCl2 + H2
Xét tỉ lệ: \(\dfrac{0,2}{1}>\dfrac{0,2}{2}\) => Zn dư, HCl hết
PTHH: Zn + 2HCl --> ZnCl2 + H2
__________0,2-------------->0,1
=> VH2 = 0,1.22,4 = 2,24(l)
b)
PTHH: 2H2 + O2 --to--> 2H2O
______0,1->0,05
=> mO2 = 0,05.22,4 = 1,12 (l)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)
a. \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)
Theo PTHH: \(n_{ZnSO_4}=n_{Zn}=0,1\left(mol\right)\)
\(\Rightarrow m_{muối}=0,1.161=16,1\left(g\right)\)
b. \(n_{H_2thu.được}=n_{Zn}=0,1\left(mol\right)\)
\(H_2+\dfrac{1}{2}O_2\underrightarrow{t^o}H_2O\)
0,1 0,05
\(V_{O_2}=0,05.22,4=1,12\left(l\right)\)
\(\Rightarrow V_{không.khí}=1,12.5=5,6\left(l\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a+b+c) PTHH: \(Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)
Ta có: \(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)=n_{ZnCl_2}=n_{H_2}\)
\(\Rightarrow\left\{{}\begin{matrix}m_{ZnCl_2}=0,1\cdot136=13,6\left(g\right)\\V_{H_2}=0,1\cdot22,4=2,24\left(l\right)\end{matrix}\right.\)
d) PTHH: \(CuO+H_2\underrightarrow{t^o}Cu+H_2O\)
Theo PTHH: \(n_{H_2}=n_{Cu}=0,1\left(mol\right)\)
\(\Rightarrow m_{Cu}=0,1\cdot64=6,4\left(g\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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: $n_{HCl}=2.n_{Zn}=0,2(mol)\Rightarrow m_{HCl}=7,3(g)$
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Fe + 2HCl --> FeCl2 + H2
b) nHCl = 0,2.1 = 0,2 (mol)
PTHH: Fe + 2HCl --> FeCl2 + H2
0,2---->0,1--->0,1
=> mFeCl2 = 0,1.127 = 12,7 (g)
c) VH2 = 0,1.22,4 = 2,24 (l)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Zn + 2HCl --> ZnCl2 + H2
b) nHCl = 0,2.1 = 0,2 (mol)
PTHH: Zn + 2HCl --> ZnCl2 + H2
0,2----->0,1---->0,1
=> mZnCl2 = 0,1.136 = 13,6 (g)
c) VH2 = 0,1.22,4 = 2,24 (l)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
\(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\\ PTHH:Zn+2HCl\rightarrow ZnCl_2+H_2\\ \left(mol\right)....0,1\rightarrow0,2.........0,1.......0,1\\ a,m_{HCl}=0,1.36,5=3,65\left(g\right)\\ b,m_{ZnCl_2}=0,1.136=13,6\left(g\right)\\c,V_{H_2} =0,1.22,4=2,24\left(l\right)\)
Bài 2:
\(n_P=\dfrac{3,1}{31}=0,1\left(mol\right)\\ PTHH:4P+5O_2\underrightarrow{t^o}2P_2O_5\\ \left(mol\right)...0,1\rightarrow0,125...0,05\\ a,m_{P_2O_5}=0,05.142=7,1\left(g\right)\\ a,V_{O_2}=0,125.22,4=2,8\left(l\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Zn + 2HCl →ZnCl2 + H2
b) nZn = 6,5/65 = 0,1 mol . Theo tỉ lệ pư => nH2 = nZn = nZnCl2 =0,1 mol <=> VH2(đktc) = 0,1.22,4 = 2,24 lít.
c) mZnCl2 = 0,1 . 136 = 13,6 gam
d) nHCl =2nZn = 0,2 mol => mHCl = 0,2.36,5= 7,3 gam
Cách 2: áp dụng định luật BTKL => mHCl = mZnCl2 + mH2 - mZn
<=> mHCl = 13,6 + 0,1.2 - 6,5 = 7,3 gam
![](https://rs.olm.vn/images/avt/0.png?1311)
2H2+O2-to>2H2O
0,2----0,1-----0,2
n H2=0,2 mol
=>m H2O=0,2.18=3,6g
=>Vkk=0,1.22,4.5=11.2l
a, \(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
PTHH: 2H2 + O2 ----to----> 2H2O
Mol: 0,2 0,1 0,2
\(m_{H_2O}=0,2.18=3,6\left(g\right)\)
b, \(V_{O_2}=0,1.22,4=2,24\left(l\right)\Rightarrow V_{kk}=2,24.5=11,2\left(l\right)\)
a, \(n_{Zn}=\dfrac{26}{65}=0,4\left(mol\right)\)
PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
Theo PT: \(n_{H_2}=n_{Zn}=0,4\left(mol\right)\Rightarrow m_{H_2}=0,4.2=0,8\left(g\right)\)
b, \(2H_2+O_2\underrightarrow{^{t^o}}2H_2O\)
Theo PT: \(n_{O_2}=\dfrac{1}{2}n_{H_2}=0,2\left(mol\right)\Rightarrow V_{O_2}=0,2.22,4=4,48\left(l\right)\)
\(\Rightarrow V_{kk}=5V_{O_2}=22,4\left(l\right)\)