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a, nZn = 97,5/65 = 1,5 (mol)
PTHH: Zn + 2HCl -> ZnCl2 + H2
nH2 = nZn = 1,5 (mol)
VH2 = 1,5 . 22,4 = 33,6 (l)
b, nFe2O3 = 120/160 = 0,75 (mol)
PTHH: Fe2O3 + 3H2 -> (t°) 2Fe + 3H2O
LTL: 0,75 > 1,5/3 => Fe2O3 dư
nFe2O3 (p/ư) = 1,5/3 = 0,5 (mol)
mFe2O3 (dư) = (0,75 - 0,5) . 160 = 40 (g)
a. \(n_{Zn}=\dfrac{97.5}{65}=1,5\left(mol\right)\)
PTHH : Zn + 2HCl -> ZnCl2 + H2
1,5 1,5
b. \(V_{H_2}=1,5.22,4=33,6\left(l\right)\)
c. \(n_{Fe_2O_3}=\dfrac{120}{160}=0,75\left(mol\right)\)
PTHH : Fe2O3 + 3H2 -> 2Fe + 3H2O
0,5 1,5
Ta thấy \(\dfrac{0.75}{1}>\dfrac{1.5}{3}\) => Fe2O3 dư
\(m_{Fe_2O_3\left(dư\right)}=\left(0,75-0,5\right).160=40\left(g\right)\)
\(V_{Fe_2O_3\left(dư\right)}=0,5.22,4=11,2\left(l\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đổi 2,016 dm3 = 2,016 l
nH2 = 2,016/22,4 = 0,09 (mol)
Gọi nFe2O3 = a (mol); nCuO = b (mol)
160a + 80b = 5,6 (g) (1)
PTHH:
Fe2O3 + 3H2 -> (t°) 2Fe + 3H2O
Mol: a ---> 3a ---> 2a ---> 3a
CuO + H2 -> (t°) Cu + H2O
Mol: b ---> b ---> b ---> b
3a + b = 0,09 (mol) (2)
Từ (1) và (2) => a = 0,02 (mol); b = 0,03 (mol)
mFe2O3 = 0,02 . 160 = 3,2 (g)
mCuO = 0,03 . 80 = 2,4 (g)
mH2O = (0,02 . 3 + 0,03) . 18 = 1,62 (g)
mFe = 2 . 0,02 . 56 = 2,24 (g)
mCu = 0,03 . 64 = 1,92 (g)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, \(Fe_3O_4+4H_2\underrightarrow{t^o}3Fe+4H_2O\)
b, \(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\)
Theo PT: \(n_{Fe_3O_4}=\dfrac{1}{3}n_{Fe}=\dfrac{1}{15}\left(mol\right)\Rightarrow m_{Fe_3O_4}=\dfrac{1}{15}.232=\dfrac{232}{15}\left(g\right)\)
c, \(n_{H_2}=\dfrac{4}{3}n_{Fe}=\dfrac{4}{15}\left(mol\right)\Rightarrow V_{H_2}=\dfrac{4}{15}.22,4=\dfrac{448}{75}\left(l\right)\)
d, \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
\(n_{Zn}=n_{H_2}=\dfrac{4}{15}\left(mol\right)\Rightarrow m_{Zn}=\dfrac{4}{15}.65=\dfrac{52}{3}\left(g\right)\)
\(n_{HCl}=2n_{H_2}=\dfrac{8}{15}\left(mol\right)\Rightarrow m_{HCl}=\dfrac{8}{15}.36,5=\dfrac{292}{15}\left(g\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(n_{H_2}=\dfrac{5,04}{22,4}=0,225\left(mol\right)\)
PTHH: CuO + H2 → Cu + H2O
Mol: x x x
PTHH: Fe2O3 + 3H2 → 2Fe + 3H2O
Mol: y 3y 2y
Ta có hpt:\(\left\{{}\begin{matrix}80x+160y=14\\x+3y=0,225\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,075\left(mol\right)\\y=0,05\left(mol\right)\end{matrix}\right.\)
\(m_{hh.kim.loại}=m_{Cu}+m_{Fe}=0,075.64+2.0,05.56=10,4\left(g\right)\)
\(n_{H_2}=\dfrac{5,04}{22,4}=0,225\left(mol\right)\)
PTHH:
\(CuO+H_2\underrightarrow{t^o}Cu+H_2O\\ Fe_2O_3+3H_2\underrightarrow{t^o}2Fe+3H_2O\)
Theo 2 pthh trên: \(n_{H_2O}=n_{H_2}=0,225\left(mol\right)\)
\(\rightarrow m_{H_2O}=0,225.18=4,05\left(g\right)\\ \rightarrow m_{H_2}=0,225.2=0,45\left(g\right)\)
Áp dụng ĐLBTKL, ta có:
\(m_{oxit\left(CuO,Fe_2O_3\right)}+m_{H_2}=m_{\text{kim loại}\left(Cu,Fe\right)}+m_{H_2O}\\ \rightarrow m_{\text{kim loại}\left(Cu,Fe\right)}=14+0,45-4,05=10,4\left(g\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, nZn = 26/65 = 0,4 (mol)
PTHH: Zn + 2HCl -> ZnCl2 + H2
nZn = nH2 = 0,4 (mol)
VH2 = 0,4 . 22,4 = 8,96 (l)
b, nFe2O3 = 16/160 = 0,1 (mol)
PTHH: Fe2O3 + 3H2 -> (t°) 2Fe + 3H2O
LTL: 0,1 < 0,4/3 => H2 dư
nFe = 0,1 . 3 = 0,3 (mol)
mFe = 0,3 . 56 = 16,8 (g)
a) \(n_{Zn}=\dfrac{26}{65}=0,4\left(mol\right)\)
PTHH: Zn + 2HCl --> ZnCl2 + H2
0,4--------------------->0,4
=> VH2 = 0,4.22,4 = 8,96 (l)
b)
\(n_{Fe_2O_3}=\dfrac{16}{160}=0,1\left(mol\right)\)
PTHH: Fe2O3 + 3H2 --to--> 2Fe + 3H2O
Xét tỉ lệ: \(\dfrac{0,1}{1}< \dfrac{0,4}{3}\) => Fe2O3 hết, H2 dư
PTHH: Fe2O3 + 3H2 --to--> 2Fe + 3H2O
0,1---------------->0,2
=> mFe = 0,2.56 = 11,2 (g)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, \(n_{Fe}=\dfrac{16,8}{56}=0,3\left(mol\right)\)
PTHH: Fe + 2HCl ---to---> FeCl2 + H2
Mol: 0,3 0,6 0,3
\(V_{H_2}=0,3.22,4=6,72\left(l\right)\)
\(m_{HCl}=0,6.36,5=21,9\left(g\right)\)
b, \(n_{CuO}=\dfrac{32}{80}=0,4\left(mol\right)\)
PTHH: H2 + CuO ---to---> Cu + H2O
Mol: 0,3 0,3
Ta có: \(\dfrac{0,3}{1}< \dfrac{0,4}{1}\) ⇒ H2 pứ hết, CuO dư
\(m_{Cu}=0,3.64=19,2\left(g\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(n_{Zn}=\dfrac{13}{65}=0,2(mol)\\ a,PTHH:Zn+2HCl\to ZnCl_2+H_2\\ b,n_{HCl}=2n_{Zn}=0,4(mol)\\ \Rightarrow m_{HCl}=0,4.36,5=14,6(g)\\ c,n_{H_2}=n_{Zn}=0,2(mol)\\ \Rightarrow V_{H_2}=0,2.22,4=4,48(l)\)
b) mHCl = 14,6 (g)
V H2 = 4,48 (l)
Giải thích các bước:
a) PTHH: Zn + 2HCl → ZnCl2 + H2↑
b) nZn = 13 : 65 = 0,2 mol
Theo PTHH: nHCl = 2.nZn = 0,4 mol
mHCl = 0,4 . 36,5 = 14,6(g)
c) nH2 = nZn = 0,2 mol
VH2 = 0,2 . 22,4 = 4,48 (l)