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\(a,4Na+O_2\xrightarrow{t^o}2Na_2O\\ b,Mg+2HCl\to MgCl_2+H_2\\ c,6NaOH+Fe_2(SO_4)_3\to 3Na_2SO_4+2Fe(OH)_3\downarrow\)
\(a.4Na+O_2-^{t^o}\rightarrow2Na_2O\\ b.Mg+2HCl\rightarrow MgCl_2+H_2\\ c.6NaOH+Fe_2\left(SO_4\right)_3\rightarrow3Na_2SO_4+2Fe\left(OH\right)_3\)
\(n_{Mg}=\dfrac{4.8}{24}=0.2\left(mol\right)\)
\(Mg+2HCl\rightarrow MgCl_2+H_2\)
\(0.2......................0.2......0.2\)
\(m_{MgCl_2}=0.2\cdot95=19\left(g\right)\)
\(V_{H_2}=0.2\cdot22.4=4.48\left(l\right)\)
a) Mg + 2HCl --> MgCl2 + H2
b) \(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\)
Mg + 2HCl --> MgCl2 + H2
0,2--------------->0,2--->0,2
=> mMgCl2 = 0,2.95=19 (g)
c) VH2 = 0,2.22,4 = 4,48(l)
\(1.2Fe+3Cl_2\overset{t^o}{--->}2FeCl_3\)
\(2.Zn+S\overset{t^o}{--->}ZnS\)
\(3.4P+5O_2\overset{t^o}{--->}2P_2O_5\)
\(4.Mg+2HCl--->MgCl_2+H_2\)
\(5.CO_2+H_2O--->H_2CO_3\)
\(6.K_2O+H_2O--->2KOH\)
\(7.4Na+O_2--->2Na_2O\)
\(8.Fe_2\left(SO_4\right)_3+3Ca\left(OH\right)_2--->2Fe\left(OH\right)_3\downarrow+3CaSO_4\)
\(9.Al_2O_3+3H_2SO_4--->Al_2\left(SO_4\right)_3+3H_2O\)
1) 2Fe+3Cl2 --to- > 2FeCl3
2)Zn+S --to- > ZnS
3) 4P+5O2 --to- > 2P2O5
4) Mg+ 3HCl ---> MgCl2 + H2
5)CO2+H2O --->H2CO3
6)K2O+H2O ----> 2KOH
7)4Na + O2 --to- > 2Na2O
8)Fe2(SO4)3 + 3Ca(OH)2 ----> 2Fe(OH)3+ 3CaSO4
9. Al2O3 + 3H2SO4 -----> Al2(SO4)3 + 3H2O
Nếu có thể thì lần sau bạn nên đăng tách từng bài ra nhé!
Bài 1:
PT: \(Mg+2HCl\rightarrow MgCl_2+H_2\)
Ta có: \(n_{Mg}=\dfrac{2,4}{24}=0,1\left(mol\right)\)
\(n_{HCl}=\dfrac{3,65}{36,5}=0,1\left(mol\right)\)
Xét tỉ lệ: \(\dfrac{0,1}{1}>\dfrac{0,1}{2}\) , ta được Mg dư.
Theo PT: \(n_{Mg\left(pư\right)}=n_{MgCl_2}=n_{H_2}=\dfrac{1}{2}n_{HCl}=0,05\left(mol\right)\)
\(\Rightarrow n_{Mg\left(dư\right)}=0,1-0,05=0,05\left(mol\right)\)
\(\Rightarrow m_{Mg\left(dư\right)}=0,05.24=1,2\left(g\right)\)
\(m_{MgCl_2}=0,05.95=4,75\left(g\right)\)
\(V_{H_2}=0,05.22,4=1,12\left(l\right)\)
Bài 2:
PT: \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
Ta có: \(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)
\(n_{H_2SO_4}=\dfrac{14,7}{98}=0,15\left(mol\right)\)
Xét tỉ lệ: \(\dfrac{0,2}{2}>\dfrac{0,15}{3}\) , ta được Al dư.
Theo PT: \(\left\{{}\begin{matrix}n_{Al\left(pư\right)}=\dfrac{2}{3}n_{H_2SO_4}=0,1\left(mol\right)\\n_{Al_2\left(SO_4\right)_3}=\dfrac{1}{3}n_{H_2SO_4}=0,05\left(mol\right)\\n_{H_2}=n_{H_2SO_4}=0,15\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow n_{Al\left(dư\right)}=0,2-0,1=0,1\left(mol\right)\)
\(\Rightarrow m_{Al\left(dư\right)}=0,1.27=2,7\left(g\right)\)
\(m_{Al_2\left(SO_4\right)_3}=0,05.342=17,1\left(g\right)\)
\(V_{H_2}=0,15.22,4=3,36\left(l\right)\)
Bài 3:
PT: \(2M+6HCl\rightarrow2MCl_3+3H_2\)
Ta có: \(n_{H_2}=\dfrac{4,704}{22,4}=0,21\left(mol\right)\)
Theo PT: \(n_M=\dfrac{2}{3}n_{H_2}=0,14\left(mol\right)\)
\(\Rightarrow M_M=\dfrac{3,78}{0,14}=27\left(g/mol\right)\)
Vậy: M là nhôm (Al).
Bài 4:
PT: \(2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\)
Theo PT: \(n_{O_2}=\dfrac{1}{2}n_{KMnO_4}=0,2\left(mol\right)\)
PT: \(4P+5O_2\underrightarrow{t^o}2P_2O_5\)
Ta có: \(n_P=\dfrac{6,2}{31}=0,2\left(mol\right)\)
Xét tỉ lệ: \(\dfrac{0,2}{4}>\dfrac{0,2}{5}\) , ta được P dư.
Theo PT: \(n_{P_2O_5}=\dfrac{2}{5}n_{O_2}=0,08\left(mol\right)\)
\(\Rightarrow m_{P_2O_5}=0,08.142=11,36\left(g\right)\)
Bạn tham khảo nhé!
a) 3Fe +2 O2 --to--> Fe3O4
b) 4P + 5O2 --to--> 2P2O5
c) Al2O3 + 3H2SO4 --> Al2(SO4)3 + 3H2O
d) Fe2O3 + 3H2SO4 --> Fe2(SO4)3 + 3H2O
e) 2Cu(NO3)2 --to--> 2CuO + 4NO2 + O2
\(a,3Fe+2O_2\xrightarrow{t^o}Fe_3O_4\\ b,4P+5O_2\xrightarrow{t^o}2P_2O_5\\ c,Al_2O_3+3H_2SO_4\to Al_2(SO_4)_3+3H_2O\\ d,Fe_2O_3+3H_2SO_4\to Fe_2(SO_4)_3+3H_2O\\ e,2Cu(NO_3)_2\xrightarrow{t^o}2CuO+4NO_2+O_2\uparrow\)
a) $4P + 5O_2 \xrightarrow{t^o} 2P_2O_5$
b) $Mg + Cl_2 \xrightarrow{t^o} MgCl_2$
c) $2Na + 2H_2O \to 2NaOH + H_2$
d) $C + O_2 \xrightarrow{t^o} CO_2$
e) $C_xH_y + (x + \dfrac{y}{4})O_2 \xrightarrow{t^o} xCO_2 + \dfrac{y}{2}H_2O$
f) $2Al + Fe_2O_3 \xrightarrow{t^o} Al_2O_3 + 2Fe$
g) $2Al + 3H_2SO_4 \to Al_2(SO_4)_3 + 3H_2$
i) $Fe_xO_y + yCO \xrightarrow{t^o} xFe + yCO_2$
k) $Fe_2O_3 + 6HCl \to 2FeCl_3 +3 H_2O$
l) $3Fe + 2O_2 \xrightarrow{t^o} Fe_3O_4$