Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(a, 2Al+3Cl_2->2AlCl_3\)
\(b,2KClO_3->2KCl+3O_2\)
\(c, 2H_3PO_4+3Ba(OH)_2->Ba_3(PO_4)_2+6H_2O\)
\(\left(a\right)2Al+3Cl_2\overset{t^o}{--->}2AlCl_3\)
\(\left(b\right)2KClO_3\xrightarrow[MnO_2]{t^o}2KCl+3O_2\uparrow\)
\(\left(c\right)2NaOH+CuSO_4--->Cu\left(OH\right)_2\downarrow+Na_2SO_4\)
\(\left(d\right)2H_3PO_4+3Ba\left(OH\right)_2--->Ba_3\left(PO_4\right)_2+6H_2O\)
Ca + H2O \(\rightarrow\) Ca(OH)2 + H2
P2O5 + 3H2O \(\rightarrow\) 2H3PO4
2Al(OH)3 + 3H2SO4 \(\rightarrow\) Al2(SO4)3 + 6H2O
3Mg + 2AlCl3 \(\rightarrow\) 2Al + 3MgCl2
2KMnO4 + 16HCl \(\rightarrow\) 2KCl + 2MnCl2 + 5Cl2 + 8H2O
2Zn + O2 \(\rightarrow\) 2ZnO
3CO + Fe2O3 \(\rightarrow\) 2Fe + 3CO2
Bài 2 :
a) \(n_{Al}=\dfrac{9,2}{27}=0,34\left(mol\right)\)
\(PTHH:2Na+2H_2O->2NaOH+H_2\)
0,17 0,17 0,17 0,17
\(V_{H_2}=0,17.22,4=3,808\left(l\right)\)
\(m_{NaOH}=0,17.40=6,8\left(g\right)\)
a, \(CuO+H_2\underrightarrow{t^o}Cu+H_2O\)
b, \(CO_2+2NaOH\rightarrow Na_2CO_3+H_2O\)
c, \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
d, \(4Al+3O_2\underrightarrow{t^o}2Al_2O_3\)
e, \(2NaOH+CuSO_4\rightarrow Na_2SO_4+Cu\left(OH\right)_2\)
f, \(Al_2O_3+2NaOH\rightarrow2NaAlO_2+H_2O\)
g, \(2Fe\left(OH\right)_3\underrightarrow{t^o}Fe_2O_3+3H_2O\)
h, \(2H_3PO_4+3Ca\left(OH\right)_2\rightarrow Ca_3\left(PO_4\right)_2+6H_2O\)
i, \(BaCl_2+2AgNO_3\rightarrow2AgCl+Ba\left(NO_3\right)_2\)
k, \(2FeO+4H_2SO_{4đ}\rightarrow Fe_2\left(SO_4\right)_3+SO_2+4H_2O\)
Bạn tham khảo nhé!
4K+O2->2K2O
2Fe(OH)3->Fe2O3+3H2O
2SO2+O2->2SO3
Cu+2AgNO3->Cu(NO3)2+2Ag
\(1,2Ba+O_2\xrightarrow{t^o}2BaO\\ 2,Zn(OH)_2+2HCl\to ZnCl_2+2H_2O\\ 3,P_2O_5+3H_2O\to 2H_3PO_4\\ 4,C_4H_8+6O_2\xrightarrow{t^o}4CO_2+4H_2O\\ 5,2NaHCO_3+H_2SO_4\to Na_2SO_4+2H_2O+2CO_2\uparrow\\ 6,\text{Sẽ có }22\text{ phân tử }CO_2{\text{ được tạo ra}}\)
\(1,4Na+O_2\rightarrow^{t^o}2Na_2O\\ 2,4P+5O_2\rightarrow^{t^o}2P_2O_5\\ 3,3Fe+2O_2\rightarrow^{t^o}Fe_3O_4\\ 4,2H_2+O_2\rightarrow^{t^o}2H_2O\\ 5,N_2+H_2⇌^{\left(t^o,xt,p\right)}NH_3\\ 6,P_2O_5+3H_2O\rightarrow2H_3PO_4\\ 7,N_2O_5+H_2O\rightarrow2HNO_3\\ 8,2Al+6HCl\rightarrow2AlCl_3+3H_2\\ 9,2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\\ 10,2Cu+O_2\rightarrow^{t^o}2CuO\\ 11,CaO+2HNO_3\rightarrow Ca\left(NO_3\right)_2+H_2O\\ 12,Al\left(OH\right)_3+3HNO_3\rightarrow Al\left(NO_3\right)_3+3H_2O\\ 13,Zn+2HCl\rightarrow ZnCl_2+H_2\)
Gọi \(\left\{{}\begin{matrix}n_{Fe}=a\left(mol\right)\\n_{Mg}=b\left(mol\right)\end{matrix}\right.\)
\(n_{CuO}=\dfrac{5}{80}=0,0625\left(mol\right)\)
PTHH:
Fe + H2SO4 ---> FeSO4 + H2
a ----------------------> a
Mg + H2SO4 ---> MgSO4 + H2
b ------------------------> b
MgSO4 + 2KOH ---> Mg(OH)2 + K2SO4
b ---------------------------> b
FeSO4 + 2KOH ---> Fe(OH)2 + K2SO4
a ---------------------------> a
4Fe(OH)2 + O2 + 2H2O --to--> 4Fe(OH)3
a ----------------------------------------> a
2Fe(OH)3 --to--> Fe2O3 + 3H2O
a --------------------> 0,5a
Mg(OH)2 --to--> MgO + H2O
b -------------------> b
2Cu + O2 --to--> 2CuO
0,0625 <------------- 0,0625
Hệ pt \(\left\{{}\begin{matrix}56a+24b+0,0625.64=20\\160.0,5a+40b=24\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,2\left(mol\right)\\b=0,2\left(mol\right)\end{matrix}\right.\)
\(\rightarrow\left\{{}\begin{matrix}m_{Fe}=0,2.56=11,2\left(g\right)\\m_{Mg}=0,2.24=4,8\left(g\right)\\m_{Cu}=0,625.64=4\left(g\right)\end{matrix}\right.\rightarrow\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{11,2}{20}=56\%\\\%m_{Mg}=\dfrac{4,8}{20}=24\%\\\%m_{Cu}=\dfrac{4}{20}=20\%\end{matrix}\right.\)
\(\left\{{}\begin{matrix}Fe:x\left(mol\right)\\Mg:y\left(mol\right)\end{matrix}\right.\)
\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)
x x x
\(Mg+H_2SO_4\rightarrow MgSO_4+H_2\)
y y y
Khí A nhận được là \(H_2\)
\(FeSO_4+2KOH\rightarrow Fe\left(OH\right)_2\downarrow+K_2SO_4\)
x x
\(MgSO_4+2KOH\rightarrow Mg\left(OH\right)_2\downarrow+K_2SO_4\)
y y
\(2Fe\left(OH\right)_3\underrightarrow{t^o}Fe_2O_3+3H_2O\)
x \(\dfrac{x}{2}\)
\(Mg\left(OH\right)_2\underrightarrow{t^o}MgO+H_2O\)
y y
\(2Cu+O_2\underrightarrow{t^o}2CuO\)
\(n_{CuO}=\dfrac{5}{80}=0,0625mol\Rightarrow m_{Cu}=4g\)
\(m_{Fe+Mg}=20-4=16g\)
\(\Rightarrow\left\{{}\begin{matrix}56x+24y=16\\80x+40y=24\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,2\\y=0,2\end{matrix}\right.\)
\(\%m_{Fe}=\dfrac{0,2\cdot56}{20}\cdot100\%=56\%\)
\(\%m_{Mg}=\dfrac{0,2\cdot24}{20}\cdot100\%=24\%\)
\(\%m_{Cu}=100\%-\left(56\%+24\%\right)=20\%\)
Câu 4:
\(a,2Mg+O_2\underrightarrow{t^o}2MgO\)
\(b,2Fe\left(OH\right)_3\underrightarrow{t^o}Fe_2O_3+3H_2O\)
\(c,2NaOH+CuSO_4\rightarrow Cu\left(OH\right)_2+Na_2SO_4\)
\(d,P_2O_5+3Ca\left(OH\right)_2\rightarrow Ca_3\left(PO_4\right)_2+3H_2O\) (thiếu O ở H2O nha)
Câu 5:
\(a,V_{O_2}=n.22,4=0,5.22,4=11,2\left(l\right)\\ b,n_{SO_2}=\dfrac{m}{M}=\dfrac{19,2}{32}=0,6\left(mol\right)\\ V_{SO_2}=n.22,4=0,6.22,4=13,44\left(l\right)\)