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a) nAl= 0,2(mol)
PTHH: 4Al + 3 O2 -to-> 2 Al2O3
nO2= 3/4 . 0,2= 0,15(mol)
=>V(O2,đktc)=0,15.22,4=3,36(l)
Vkk(đktc)=5.V(O2,đktc)=3,36.5=16,8(l)
b) nAl=0,2(mol)
nO2=0,4(mol)
Ta có: 0,2/4 < 0,4/3
=> Al hết, O2 dư, tính theo nAl.
- Sau phản ứng có O2(dư) và Al2O3
nAl2O3= nAl/2= 0,2/2=0,1(mol)
nO2(dư)= 0,4- 0,2. 3/4=0,25(mol)
1.
a, \(Natri+sulfuric\:acid\rightarrow Natri\text{ }sulfat+Hidro\)
b, PTHH: \(2Na+H_2SO_4\rightarrow Na_2SO_4+H_2\uparrow\)
Câu 1:
\(a,\) sodium + sulfuric acid ---> sodium sulfate + hydrogen
\(b,2Na+H_2SO_4\to Na_2SO_4+H_2\)
Câu 2:
\(a,\) iron sulfate + barium chloride -----> barium sulfate + iron chloride
\(b,Fe_2\left(SO_4\right)_3+3BaCl_2\to3BaSO_4\downarrow+2FeCl_3\)
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1 đơn vị cacbon hay ghi tắt là 1đ.v.c bằng:
(1,9926.10-23)/12 (g)
Câu 1:
Cacbon: C
Sắt: Fe
Oxi: O2
Magie: Mg
Photpho : P
Canxi: Ca
Hidro: H2
Natri: Na
Lưu huỳnh: S
Clo: Cl2
Bài 1:
\(a,Magnesium+Oxygen\xrightarrow{t^o}Magnesium Oxide\\ b,m_{Mg}+m_{O_2}=m_{MgO}\\ c,m_{O_2}=m_{MgO}-m_{Mg}=15-9=6(g)\)
Bài 2:
\(a,Sulfur+Oxygen\xrightarrow{t^o}\text {Sulfur dioxide}\\ b,m_{S}+m_{O_2}=m_{SO_2}\\ c,m_{O_2}=m_{SO_2}-m_{S}=6,4-3,2=3,2(g)\)
Bài 3:
\(a,zinc+\text{hydrochloric acid}\to \text {zinc chloride}+hydrogen\\ b,m_{Zn}+m_{HCl}=m_{ZnCl_2}+m_{H_2}\\ c,m_{HCl}=m_{ZnCl_2}+m_{H_2}-m_{Zn}=13,6+0,2-6,5=7,3(g)\)
Cảm ơn