Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) Gọi số mol Al, Zn là 2a, a (mol)
PTHH: 4Al + 3O2 --to--> 2Al2O3
2a-->1,5a---------->a
2Zn + O2 --to--> 2ZnO
a---->0,5a------->a
=> \(102a+81a=18,3\)
=> a = 0,1 (mol)
=> \(\left\{{}\begin{matrix}\%m_{Al}=\dfrac{0,2.27}{0,2.27+0,1.65}.100\%=45,378\%\\\%m_{Zn}=\dfrac{0,1.65}{0,2.27+0,1.65}.100\%=54,622\%\end{matrix}\right.\)
b) \(n_{O_2}=1,5a+0,5a=0,2\left(mol\right)\)
=> \(V_{O_2}=0,2.22,4=4,48\left(l\right)\)
a)
Theo ĐLBTKL: \(m_{Fe\left(bđ\right)}+m_{O_2}=m_X\)
=> \(m_{O_2}=26,4-20=6,4\left(g\right)\)
=> \(n_{O_2}=\dfrac{6,4}{32}=0,2\left(mol\right)\Rightarrow V=0,2.22,4=4,48\left(l\right)\)
b)
PTHH: 3Fe + 2O2 --to--> Fe3O4
0,2------->0,1
=> \(\%m_{Fe_3O_4}=\dfrac{0,1.232}{26,4}.100\%=87,88\%\)
c)
- Nếu dùng KClO3
PTHH: 2KClO3 --to--> 2KCl + 3O2
\(\dfrac{0,4}{3}\)<-----------------0,2
=> \(m_{KClO_3}=\dfrac{0,4}{3}.122,5=\dfrac{49}{3}\left(g\right)\)
- Nếu dùng KMnO4:
PTHH: 2KMnO4 --to--> K2MnO4 + MnO2 + O2
0,4<--------------------------------0,2
=> \(m_{KMnO_4}=0,4.158=63,2\left(g\right)\)
a. Ag không phản ứng nên ta có PTHH: \(2Mg+O_2\rightarrow^{t^o}2MgO\)
\(\rightarrow m_{O_2}=m_{hh}-m_{\mu\text{ối}}=18,8-15,6=3,2g\)
\(\rightarrow n_{O_2}=\frac{3,2}{32}=0,1mol\)
b. \(\rightarrow V_{O_2}=n.22,4=22,4.0,1=2,24l\)
\(\rightarrow V_{kk}=4,48.5=11,2l\)
c. Có \(n_{Mg}=2n_{O_2}=0,2l\)
\(\rightarrow m_{Mg}=0,2.24=4,8g\)
\(\rightarrow\%m_{Mg}=\frac{4,8.100}{15,6}\approx30,77\%\)
\(\rightarrow\%m_{Ag}=100\%-30,77\%=69,23\%\)
\(BTKL:\)
\(m_A+m_{O_2}=m_B\)
\(\Rightarrow m_{O_2}=m_B-m_A=32-22.4=9.6\left(g\right)\)
\(n_{O_2}=\dfrac{9.6}{32}=0.3\left(mol\right)\)
\(V_{O_2}=0.3\cdot22.4=6.72\left(l\right)\)
a) PTHH: \(2CO+O_2\underrightarrow{t^o}2CO_2\) (1)
\(4H_2+O_2\underrightarrow{t^o}2H_2O\) (2)
b) Ta có: \(\left\{{}\begin{matrix}\Sigma n_{O_2}=\dfrac{9,6}{32}=0,3\left(mol\right)\\n_{CO_2}=\dfrac{8,8}{44}=0,2\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}n_{O_2\left(1\right)}=0,1mol\\n_{O_2\left(2\right)}=0,2mol\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}m_{CO}=0,1\cdot28=2,8\left(g\right)\\m_{H_2}=0,2\cdot2=0,4\left(g\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\%m_{CO}=\dfrac{2,8}{2,8+0,4}\cdot100\%=87,5\%\\\%m_{H_2}=12,5\%\end{matrix}\right.\)
c) PTHH: \(2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\uparrow\)
Theo PTHH: \(n_{KMnO_4}=2n_{O_2}=0,6mol\)
\(\Rightarrow m_{KMnO_4}=0,6\cdot158=94,8\left(g\right)\)
\(4Al + 3O_2 \xrightarrow{t^o} 2Al_2O_3\\ 3Fe + 2O_2 \xrightarrow{t^o} Fe_3O_4\\ Fe_3O_4 + 4H_2 \xrightarrow{t^o} 3Fe + 4H_2O\\ Ta\ có :\\ m_O = m_B - m_{hh} = 5,4 - 4,44 = 0,96(mol)\\ n_O = \dfrac{0,96}{32} = 0,03(mol)\\ \Rightarrow n_{Al_2O_3}= \dfrac{1}{3}n_O = 0,01(mol)\\ \Rightarrow n_{Al} = 2n_{Al_2O_3} = 0,02(mol)\\ m_{Al} = 0,02.54 = 1,08(gam)\\ m_{Fe} = 4,44 - 1,08 = 3,36(gam)\)
a) PTTH: \(2H_2+O_2\rightarrow2H_2O\)
\(C_2H_2+\dfrac{5}{2}O_2\underrightarrow{t^o}2CO_2+H_2O\)
b) Ta có: \(\left\{{}\begin{matrix}\overline{M}_{hhkhí}=0,5\cdot28=14\\n_{hhkhí}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\end{matrix}\right.\)
Theo phương pháp đường chéo, ta có: \(\dfrac{n_{H_2}}{n_{C_2H_2}}=\dfrac{12}{12}=1\)
\(\Rightarrow n_{H_2}=n_{C_2H_2}=0,3\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}\%V_{H_2}=\dfrac{0,3}{0,6}\cdot100\%=50\%\\\%V_{C_2H_2}=50\%\\\%m_{H_2}=\dfrac{0,3\cdot2}{5,6}\cdot100\%\approx10,71\%\\\%m_{C_2H_4}=89,29\%\end{matrix}\right.\)
a) \(n_{SO_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)
PTHH: S + O2 --to--> SO2
0,5<-0,5<------0,5
=> mS = 0,5.32 = 16(g)
=> \(\left\{{}\begin{matrix}\%m_S=\dfrac{16}{22,2}.100\%=72,07\%\\\%m_P=\dfrac{22,2-16}{22,2}.100\%=27,93\%\end{matrix}\right.\)
b) \(n_P=\dfrac{22,2-16}{31}=0,2\left(mol\right)\)
PTHH: 4P + 5O2 --to--> 2P2O5
0,2-->0,25----->0,1
=> \(V_{O_2}=0,25.22,4=5,6\left(l\right)\)
c)
PTHH: 2KClO3 --to--> 2KCl + 3O2
0,5<-------------------0,75
=> \(m_{KClO_3}=0,5.122,5=61,25\left(g\right)\)
a) PTHH:
\(S+O_2\rightarrow\left(t^o\right)SO_2\\ 4P+5O_2\rightarrow\left(t^o\right)2P_2O_5\)
- Chất khí mùi hắc là SO2
- Chất rắn sau phản ứng có m(g) là P2O5
Đặt: nS=a(mol); nP=b(mol) (a,b>0) (nguyên, dương)
\(\Rightarrow\left\{{}\begin{matrix}32a+31b=22,2\\22,4a=11,2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,5\\b=0,2\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}\%m_S=\dfrac{0,5.32}{22,2}.100\approx72,072\%\\\%m_P\approx100\%-72,072\%\approx27,928\%\end{matrix}\right.\)
b)
\(n_{O_2}=a+\dfrac{5}{4}b=0,5+\dfrac{5}{4}.0,2=0,75\left(mol\right)\\ \Rightarrow V_{O_2\left(đktc\right)}=0,75.22,4=16,8\left(l\right)\)
c)
\(2KClO_3\rightarrow\left(t^o\right)2KCl+3O_2\\ n_{KClO_3}=\dfrac{2}{3}.n_{O_2}=\dfrac{2.0,75}{3}=0,5\left(mol\right)\\ \Rightarrow m_{KClO_3}=122,5.0,5=61,25\left(g\right)\)