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PT: \(FeO+H_2\underrightarrow{t^o}Fe+H_2O\)
Ta có: \(n_{FeO}=\dfrac{7,2}{72}=0,1\left(mol\right)\)
a, Theo PT: \(n_{Fe}=n_{FeO}=0,1\left(mol\right)\)
\(\Rightarrow m_{Fe}=0,1.56=5,6\left(g\right)\)
b, Theo PT: \(n_{H_2}=n_{FeO}=0,1\left(mol\right)\)
\(\Rightarrow V_{H_2}=0,1.22,4=2,24\left(l\right)\)
c, PT: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
THeo PT: \(n_{Al}=\dfrac{2}{3}n_{H_2}=\dfrac{1}{15}\left(mol\right)\)
\(\Rightarrow m_{Al}=\dfrac{1}{15}.27=1,8\left(g\right)\)
a.b.
\(n_{Fe}=\dfrac{2,8}{56}=0,05mol\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
0,05 0,1 0,05 ( mol )
\(V_{H_2}=0,05.22,4=1,12l\)
\(m_{HCl}=0,1.36,5=3,65g\)
c.
\(CuO+H_2\rightarrow\left(t^o\right)Cu+H_2O\)
0,05 0,05 ( mol )
\(m_{Cu}=0,05.64=3,2g\)
a, nZn = 26/65 = 0,4 (mol)
PTHH: Zn + 2HCl -> ZnCl2 + H2
nZn = nH2 = 0,4 (mol)
VH2 = 0,4 . 22,4 = 8,96 (l)
b, nFe2O3 = 16/160 = 0,1 (mol)
PTHH: Fe2O3 + 3H2 -> (t°) 2Fe + 3H2O
LTL: 0,1 < 0,4/3 => H2 dư
nFe = 0,1 . 3 = 0,3 (mol)
mFe = 0,3 . 56 = 16,8 (g)
a) \(n_{Zn}=\dfrac{26}{65}=0,4\left(mol\right)\)
PTHH: Zn + 2HCl --> ZnCl2 + H2
0,4--------------------->0,4
=> VH2 = 0,4.22,4 = 8,96 (l)
b)
\(n_{Fe_2O_3}=\dfrac{16}{160}=0,1\left(mol\right)\)
PTHH: Fe2O3 + 3H2 --to--> 2Fe + 3H2O
Xét tỉ lệ: \(\dfrac{0,1}{1}< \dfrac{0,4}{3}\) => Fe2O3 hết, H2 dư
PTHH: Fe2O3 + 3H2 --to--> 2Fe + 3H2O
0,1---------------->0,2
=> mFe = 0,2.56 = 11,2 (g)
a,\(n_{Mg}=\dfrac{9,6}{24}=0,4\left(mol\right);n_{H_2SO_4}=1,5.0,2=0,3\left(mol\right)\)
PTHH: Mg + H2SO4 → MgSO4 + H2
Mol: 0,3 0,3 0,3
Ta có: \(\dfrac{0,4}{1}>\dfrac{0,3}{1}\) ⇒ Mg dư, H2SO4 pứ hết
\(m_{MgSO_4}=0,3.120=36\left(g\right)\)
b,\(V_{H_2}=0,3.22,4=6,72\left(l\right)\)
c, \(n_{Fe_2O_3}=\dfrac{6,4}{160}=0,04\left(mol\right)\)
PTHH: 3H2 + Fe2O3 → 2Fe + 3H2O
Mol: 0,04 0,08
Ta có: \(\dfrac{0,3}{3}>\dfrac{0,04}{1}\) ⇒ H2 dư, Fe2O3 pứ hết
\(\Rightarrow m_{Fe}=0,08.56=4,48\left(g\right)\)
\(n_{Mg}=\dfrac{3,6}{24}=0,15\left(mol\right)\)
\(PTHH:Mg+2HCl\rightarrow MgCl_2+H_2\)
\(\left(mol\right)\) \(0,15\) \(0,3\) \(0,15\) \(0,15\)
\(a.V_{H_2}=0,15.22,4=3,36\left(l\right)\\ b.m_{MgCl_2}=95.0,15=14,25\left(g\right)\\ c.\\ PTHH:CuO+H_2\underrightarrow{t^o}Cu+H_2O\)
\(\left(mol\right)\) \(0,15\) \(0,15\) \(0,15\)
\(m_{Cu}=0,15.64=9,6\left(g\right)\)
`n_[Fe]=[5,6]/56=0,1(mol)`
`n_[HCl]=[10,95]/[36,5]=0,3(mol)`
`Fe + 2HCl -> FeCl_2 + H_2`
`0,1` `0,2` `0,1` `(mol)`
Ta có:`[0,1]/1 < [0,3]/2`
`=>HCl` dư
`a)V_[H_2]=0,1.22,4=2,24(l)`
`b)`
`H_2 + CuO` $\xrightarrow{t^o}$ `Cu + H_2 O`
`0,1` `0,1` `(mol)`
`=>m_[Cu]=0,1.64=6,4(g)`
\(a,n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\\ n_{HCl}=\dfrac{10,95}{36,5}=0,3\left(mol\right)\)
PTHH: \(Fe+2HCl\rightarrow FeCl_2+H_2\)
bđ 0,1 0,3
pư 0,1 0,2
spư 0 0,1 0,1 0,1
\(\rightarrow V_{H_2}=0,1.22,4=2,24\left(l\right)\)
b, PTHH: \(H_2+CuO\xrightarrow[]{t^o}Cu+H_2O\)
0,1------------>0,1
\(\rightarrow m_{Cu}=0,1.64=6,4\left(g\right)\)
PT: \(Fe_2O_3+3H_2\underrightarrow{t^o}2Fe+3H_2O\)
Ta có: \(n_{Fe_2O_3}=\dfrac{16}{160}=0,1\left(mol\right)\)
a, \(n_{Fe}=2n_{Fe_2O_3}=0,2\left(mol\right)\)
\(\Rightarrow m_{Fe}=0,2.56=11,2\left(g\right)\)
b, \(n_{H_2}=3n_{Fe_2O_3}=0,3\left(mol\right)\)
\(\Rightarrow V_{H_2}=0,3.22,4=6,72\left(l\right)\)
c, PT: \(Mg+2HCl\rightarrow MgCl_2+H_2\)
Theo PT: \(n_{Mg}=n_{H_2}=0,3\left(mol\right)\)
\(\Rightarrow m_{Mg}=0,3.24=7,2\left(g\right)\)
thank you