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a) \(n_{Mg}=\dfrac{2,4}{24}=0,1\left(mol\right)\)
PTHH: Mg + 2HCl --> MgCl2 + H2
_____0,1--->0,2------->0,1---->0,1
=> mHCl = 0,2.36,5 = 7,3(g)
b) mMgCl2 = 0,1.95 = 9,5 (g)
c) VH2 = 0,1.22,4 = 2,24(l)
200ml = 0,2l
\(n_{HCl}=0,5.0,2=0,1\left(mol\right)\)
a) Pt : \(Mg+2HCl\rightarrow MgCl_2+H_2|\)
1 2 1 1
0,05 0,1 0,1 0,05
b) \(n_{Mg}=\dfrac{0,1.2}{1}=0,05\left(mol\right)\)
⇒ \(m_{Mg}=0,05.24=1,2\left(g\right)\)
c) \(n_{H2}=\dfrac{0,1.1}{2}=0,05\left(mol\right)\)
\(V_{H2\left(dktc\right)}=0,05.22,4=1,12\left(l\right)\)
d) \(n_{MgCl2}=\dfrac{0,05.1}{1}=0,05\left(mol\right)\)
\(C_{M_{MgCl2}}=\dfrac{0,05}{0,2}=0,25\left(M\right)\)
Chúc bạn học tốt
a) \(n_{Mg}=\dfrac{2,4}{24}=0,1\left(mol\right)\)
PTHH: Mg + 2HCl --> MgCl2 + H2
_____0,1--->0,2------->0,1---->0,1
=> mHCl = 0,2.36,5 = 7,3(g)
b) mMgCl2 = 0,1.95 = 9,5 (g)
c) VH2 = 0,1.22,4 = 2,24(l)
Fe + 2HCl -> FeCl2 + H2
nFe = 5,6/56 = 0,1 mol
=>nH2 = 0,1 mol
=> VH2= 0,1*22,4= 2,24 lít
\(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)
PTHH: Fe + 2HCl ---> FeCl2 + H2
0,1-->0,2------------------>0,1
=> \(\left\{{}\begin{matrix}m_{ddHCl}=\dfrac{0,2.36,5}{15\%}=\dfrac{146}{3}\left(g\right)\\V_{H_2}=0,1.22,4=4,48\left(l\right)\end{matrix}\right.\)
\(n_{Mg}=\dfrac{2.4}{24}=0.1\left(mol\right)\)
\(Mg+2HCl\rightarrow MgCl_2+H_2\)
\(0.1....................................0.1\)
\(V_{H_2}=0.1\cdot22.4=2.24\left(l\right)\)
\(2H_2+O_2\underrightarrow{^{t^0}}2H_2O\)
\(0.1.....0.05\)
\(m_{O_2\left(dư\right)}=\left(0.5-0.05\right)\cdot32=14.4\left(g\right)\)
a)
$Mg + 2HCl \to MgCl_2 + H_2$
Theo PTHH :
$n_{MgCl_2} = n_{H_2} = \dfrac{5,6}{22,4} = 0,25(mol)$
$m_{MgCl_2} = 0,25.95 = 23,75(gam)$
b)
$n_{HCl} = 2n_{H_2} = 0,5(mol)$
$m_{dd\ HCl} = \dfrac{0,5.36,5}{7,3\%} = 250(gam)$
c)
$2K + 2H_2O \to 2KOH + H_2$
$n_K = 2n_{H_2} = 0,5(mol)$
$m_K = 0,5.39 = 19,5(gam)$
a)
\(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\)
PTHH: Fe + H2SO4 --> FeSO4 + H2
0,2----------------------->0,2
=> VH2 = 0,2.22,4 = 4,48 (l)
b)
PTHH: 2H2 + O2 --to--> 2H2O
0,2-->0,1
PTHH: 2KMnO4 --to--> K2MnO4+ MnO2 + O2
0,2<------------------------------0,1
=> \(m_{KMnO_4}=0,2.158=31,6\left(g\right)\)
c) \(n_{O_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
PTHH: 3Fe + 2O2 --to--> Fe3O4
Xét tỉ lệ: \(\dfrac{0,2}{3}>\dfrac{0,1}{2}\) => Fe dư
a.\(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\)
\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\uparrow\)
0,2 0,2 ( mol )
\(V_{H_2}=0,2.22,4=4,48\left(l\right)\)
b.\(2H_2+O_2\rightarrow\left(t^o\right)2H_2O\)
0,2 0,1 ( mol )
\(2KMnO_4\rightarrow\left(t^o\right)K_2MnO_4+MnO_2+O_2\uparrow\)
0,2 0,1 ( mol )
\(m_{KMnO_4}=0,2.158=31,6\left(g\right)\)
c.\(n_{O_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
\(3Fe+2O_2\rightarrow\left(t^o\right)Fe_3O_4\)
Xét: \(\dfrac{0,2}{3}\) > \(\dfrac{0,1}{2}\) ( mol )
--> Sắt không cháy hết
a) nFe=0,1(mol); nHCl=0,4(mol)
PTHH: Fe + 2 HCl -> FeCl2 + H2
Ta có: 0,1/1 < 0,4/2
=> Fe hết, HCl dư, tish theo nFe.
b) nH2=nFeCl2=Fe=0,1(mol)
=> V(H2,đktc)=0,1.22,4=2,24(l)
c) mFeCl2=127.0,1=12,7(g)
a) nFe=0,1(mol); nHCl=0,4(mol) PTHH: Fe + 2 HCl -> FeCl2 + H2 Ta có: 0,1/1 < 0,4/2 => Fe hết, HCl dư, tish theo nFe. b) nH2=nFeCl2=Fe=0,1(mol) => V(H2,đktc)=0,1.22,4=2,24(l) c) mFeCl2=127.0,1=12,7(g)
a, \(n_{Mg}=\dfrac{3,6}{24}=0,15\left(mol\right)\)
PT: \(Mg+H_2SO_4\rightarrow MgSO_4+H_2\)
Theo PT: \(n_{H_2}=n_{Mg}=0,15\left(mol\right)\Rightarrow V_{H_2}=0,15.22,4=3,36\left(l\right)\)
b, \(n_{H_2SO_4}=n_{Mg}=0,15\left(mol\right)\Rightarrow m_{H_2SO_4}=0,15.98=14,7\left(g\right)\)
c, \(n_{Fe_3O_4}=\dfrac{13,92}{232}=0,06\left(mol\right)\)
PT: \(Fe_3O_4+4H_2\underrightarrow{t^o}3Fe+4H_2O\)
Xét tỉ lệ: \(\dfrac{0,06}{1}>\dfrac{0,15}{4}\), ta được Fe3O4 dư.
Theo PT: \(n_{Fe}=\dfrac{3}{4}n_{H_2}=0,1125\left(mol\right)\Rightarrow m_{Fe}=0,1125.56=6,3\left(g\right)\)
Câu 2 :
$a) Mg + 2HCl \to MgCl_2 + H_2$
$n_{HCl} = 0,2.1 = 0,2(mol)$
Theo PTHH : $n_{H_2} = n_{MgCl_2} = \dfrac{1}{2}n_{HCl} = 0,1(mol)$
$V_{H_2} = 0,1.22,4 = 2,24(lít)$
$b) m_{MgCl_2} = 0,1.95 = 9,5(gam)$
Câu 3 :
a) $Fe + H_2SO_4 \to FeSO_4 + H_2$
Theo PTHH :
$n_{H_2SO_4} = n_{H_2} = n_{Fe} = \dfrac{5,6}{56} = 0,1(mol)$
$C\%_{H_2SO_4} = \dfrac{0,1.98}{500}.100\% = 1,96\%$
b)
$V_{H_2} = 0,1.22,4 = 2,24(lít)$
c)
$2H_2 + O_2 \xrightarrow{t^o} 2H_2o$
$V_{O_2} = \dfrac{1}{2}V_{H_2} = 1,12(lít)$