Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Làm gộp cả phần a và b
Ta có: \(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}n_{O_2}=0,15mol\\n_{Al_2O_3}=0,1mol\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}V_{O_2}=0,15\cdot22,4=3,36\left(l\right)\\m_{Al_2O_3}=0,1\cdot102=10,2\left(g\right)\end{matrix}\right.\)
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)
b) V(kk,đktc)=3,36.5=16,8(l)
a) \(4Al + 3O_2 \xrightarrow{t^o} 2Al_2O_3\)
b)
\(n_{Al} = \dfrac{21,6}{27} = 0,8(mol)\)
Theo PTHH :
\(n_{Al_2O_3} = \dfrac{1}{2}n_{Al} = 0,4(mol)\\ \Rightarrow m_{Al_2O_3} = 0,4.102 = 40,8(gam)\)
c)
\(n_{O_2} = \dfrac{3}{4}n_{Al} = 0,6(mol)\\ \Rightarrow V_{O_2} = 0,6.22,4 = 13,44(lít)\\ \Rightarrow V_{không\ khí} = 5V_{O_2} = 13,44.5 = 67,2(lít)\)
nAl=16,2/27= 0,6(mol)
a) PTHH: 4 Al +3 O2 -to-> 2 Al2O3
nO2= 3/4 . nAl=3/4 . 0,6= 0,45(mol)
=> V(O2,đktc)=0,45 x 22,4=10,08(l)
b) nAl2O3= nAl/2=0,6/2=0,3(mol)
=>mAl2O3=102. 0,3= 30,6(g)
c) 2KMnO4 -to-> K2MnO4 + MnO2 + O2
nKMnO4= 2.nO2=2. 0,45=0,9(mol)
=>mKMnO4= 158 x 0,9= 142,2(g)
Bạn tách ra từng câu nhé!
Bài 3.
\(n_{Fe}=\dfrac{m_{Fe}}{M_{Fe}}=\dfrac{36}{56}=0,6428mol\)
\(3Fe+2O_2\rightarrow\left(t^o\right)Fe_3O_4\)
0,6428 ----- 0,4285 ( mol )
\(2KMnO_4\rightarrow\left(t^o\right)K_2MnO_4+MnO_2+O_2\)
0,857 0,4285 ( mol )
\(m_{KMnO_4}=n_{KMnO_4}.M_{KMnO_4}=0,857.158=135,406g\)
Bài 4.
a.\(n_{Al_2O_3}=\dfrac{m_{Al_2O_3}}{M_{Al_2O_3}}=\dfrac{51}{102}=0,5mol\)
\(4Al+3O_2\rightarrow\left(t^o\right)2Al_2O_3\)
1 0,75 0,5 ( mol )
\(m_{Al}=n_{Al}.M_{Al}=1.27=27g\)
\(V_{O_2}=n_{O_2}.22,4=0,75.22,4=16,8l\)
b.\(2KMnO_4\rightarrow\left(t^o\right)K_2MnO_4+MnO_2+O_2\)
1,5 0,75 ( mol )
\(m_{KMnO_4}=n_{KMnO_4}.M_{KMnO_4}=1,5.158=237g\)
\(2KClO_3\rightarrow\left(t^o\right)2KCl+3O_2\)
0,5 0,75 ( mol )
\(m_{KClO_3}=n_{KClO_3}.M_{KClO_3}=0,5.122,5=61,25g\)
a, \(4Al+3O_2\underrightarrow{t^o}2Al_2O_3\)
b, \(n_{Al_2O_3}=\dfrac{20,4}{102}=0,2\left(mol\right)\)
Theo PT: \(n_{O_2}=\dfrac{3}{2}n_{Al_2O_3}=0,3\left(mol\right)\Rightarrow V_{O_2}=0,3.22,4=6,72\left(l\right)\)
c, \(V_{kk}=\dfrac{V_{O_2}}{20\%}=33,6\left(l\right)\)
7. Ta có: nZn = \(\dfrac{13}{65}=0,2\left(mol\right)\)
\(n_{O_2}=\dfrac{32}{32}=1\left(mol\right)\)
PTHH: 2Zn + O2 ---to---> 2ZnO
Ta thấy: \(\dfrac{0,2}{2}< \dfrac{1}{1}\)
=> Oxi dư
Theo PT: nZnO = nZn = 0,2(mol)
=> mZnO = 81.0,2 = 16,2(g)
8. Ta có: nAl = \(\dfrac{21,6}{27}=0,8\left(mol\right)\)
\(n_{O_2}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\)
PTHH: 4Al + 3O2 ---to---> 2Al2O3.
Ta thấy: \(\dfrac{0,8}{4}=\dfrac{0,6}{3}\)
Vậy không có chất dư.
Theo PT: \(n_{Al_2O_3}=\dfrac{1}{2}.n_{Al}=\dfrac{1}{2}.0,8=0,4\left(mol\right)\)
=> \(m_{Al_2O_3}=0,4.102=40,8\left(g\right)\)
\(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\\ 4Al+3O_2\rightarrow\left(t^o\right)2Al_2O_3\\ n_{O_2}=\dfrac{3}{4}.0,2=0,15\left(mol\right)\\ V_{O_2\left(đktc\right)}=0,15.22,4=3,36\left(l\right)\\ V_{kk\left(đktc\right)}=5.3,36=16,8\left(l\right)\\ b,n_{Al_2O_3}=\dfrac{0,2}{2}=0,1\left(mol\right)\\ m_{Al_2O_3}=0,1.102=10,2\left(g\right)\)