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![](https://rs.olm.vn/images/avt/0.png?1311)
a. PTHH : Fe + HCl -> FeCl2 + H2
b) \(n_{Fe}=\dfrac{28}{56}=0,5\left(mol\right)\)
\(n_{FeCl_2}=\dfrac{0,5}{2}=0,25\left(mol\right)\\ m_{FeCl_2}=0,25.127=31,75\left(g\right)\)
c) \(n_{H_2}=\dfrac{0,5}{1}=0,5\left(mol\right)\\ V_{H_2}=0,5.22,4=11,2\left(l\right)\)
Mà thể tích tối đa là 10 l -> quả bóng k chứa được hết lượng H2 thoát ra ngoài
![](https://rs.olm.vn/images/avt/0.png?1311)
a, 2Mg + O2 \(\underrightarrow{t^o}\) 2MgO
b, \(n_{Mg}=\dfrac{4,8}{24}=0,2mol\)
\(n_{O_2}=\dfrac{0,2}{2}=0,1mol\)
\(m_{O_2}=0,1.32=3,2g\)
\(V_{O_2}=0,1.22,4=2,24l\)
c, Cách 1:
\(Theo.ĐLBTKL,ta.có:\\ m_{Mg}+m_{O_2}=m_{MgO}\)
\(\Rightarrow m_{MgO}=4,8+3,2=8g\)
Cách 2:
\(n_{MgO}=\dfrac{0,2.2}{2}=0,2mol\)
\(\Rightarrow m_{MgO}=0,2.40=8g\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(n_{Fe}=\dfrac{m_{Fe}}{M_{Fe}}=\dfrac{28}{56}=0,5\left(mol\right)\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
0,5-------1---------0,5------0,5
b) \(V_{H_2}=n_{H_2}.22,4=0,5.22,4=11,2\left(l\right)\)
c) \(H_2+CuO\rightarrow Cu+H_2O\)
0,5-----0,5------0,5----0,5
Khối lượng đồng tạo thành: \(m_{Cu}=n_{Cu}.64=0,5.64=32\left(g\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(n_{Fe}=\dfrac{28}{56}=0,5\left(mol\right)\)
PTHH: `Fe + 2HCl -> FeCl_2 + H_2`
0,5-------------------------->0,5`
b) `V_{H_2} = 0,5.22,4 = 11,2 (l)`
c) PTHH: \(CuO+H_2\xrightarrow[]{t^o}Cu+H_2O\)
0,5---->0,5
`=> m_{Cu} = 0,5.64 = 32 (g)`
\(Fe+2HCl\underrightarrow{t^o}FeCl_2+H_2\)
\(1mol\) \(1mol\)
\(0,5mol\) \(0,5mol\)
\(n_{Fe}=\dfrac{m}{M}=\dfrac{28}{56}=0,5\left(mol\right)\)
\(V_{H_2}=n.22,4=0,5.22,4=11,2\left(l\right)\)
\(H_2+CuO\underrightarrow{t^o}Cu+H_2O\)
\(1mol\) \(1mol\)
\(0,5mol\) \(0,5mol\)
\(m_{Cu}=n.M=0,5.64=32\left(g\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(n_{Fe}=\dfrac{28}{56}=0,5mol\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
0,5 1 0,5
\(V_{H_2}=0,5\cdot22,4=11,2l\)
\(m_{HCl}=1\cdot36,5=36,5g\)
![](https://rs.olm.vn/images/avt/0.png?1311)
BTKL: \(m_{Fe}+m_{HCl}=m_{muối}+m_{H_2}\)
\(\Rightarrow m_{H_2}=5,6+7,3-12,7=0,2\left(g\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
nMg = 3,6 : 24 = 0,15 (mol)
pthh : Mg + 2HCl --> MgCl2 + H2
0,15-----------> 0,15 --->0,15 (mol)
mMgCl2 = 0,15 . 95 = 14,25 (mol)
VH2 (đkc)= 0,15. 24,79 = 3,718(l)
\(n_{Mg}=\dfrac{3,6}{24}=0,15mol\)
\(Mg+2HCl\rightarrow MgCl_2+H_2\)
0,15 0,3 0,15 0,15
\(m_{MgCl_2}=0,15\cdot95=14,25g\)
\(V_{H_2}=0,15\cdot22,4=3,36l\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(n_{Fe}=\dfrac{m}{M}=\dfrac{28}{56}=0,5\left(mol\right)\\ PTHH:Fe+2HCl->FeCl_2+H_2\)
ti le 1 : 2 : 1 : 1
n(mol) 0,5-->1--------->0,5------>0,5
\(m_{FeCl_2}=n\cdot M=0,5\cdot\left(56+35,5\cdot2\right)=63,5\left(g\right)\\ V_{H_2\left(dktc\right)}=n\cdot22,4=0,5\cdot22,4=11,2\left(l\right)\)
\(n_{Fe}=\dfrac{28}{56}=0,5mol\)
a, PTHH : \(Fe + 2HCl-> FECl_2+ H_2↑\)
0,5 0,5 0,5 (mol)
b/. Theo phương trình, ta có:
\(n_{FeCl_2}=n_{Fe}=0,5 mol\)
\(m_{FeCl_2}=0,5.127=63,5g\)
c, Thông cảm không biết làm
\(a,PTHH\left(1\right):Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)
\(b,n_{Fe}=\dfrac{m}{M}=\dfrac{28}{56}=0,5\left(mol\right)\\ Theo.PTHH\left(1\right):n_{FeCl_2}=n_{Fe}=0,5\left(mol\right)\\ m_{FeCl_2}=n.M=0,5.91,5=45,75\left(g\right)\)
\(c,PTHH\left(2\right):2Mg+O_2\underrightarrow{t^o}2MgO\\ n_{Mg}=\dfrac{m}{M}=\dfrac{4,8}{24}=0,2\left(mol\right)\\ Theo.PTHH\left(2\right):n_{O_2}=2.n_{Mg}=0,2.2=0,4\left(mol\right)\\ V_{O_2\left(đktc\right)}=n.22,4=0,4.22,4=8,96\left(l\right)\)
(c lỗi đề à có oxi chứ ko có hidro nên mik thay bằng oxi nha)