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`a)`
`Fe + H_2 SO_4 -> FeSO_4 + H_2`
`0,4` `0,4` `0,4` `(mol)`
`n_[Fe]=[22,4]/56=0,4(mol)`
`b)m_[FeSO_4]=0,4.152=60,8(g)`
`c)V_[H_2]=0,4.22,4=8,96(l)`
\(Zn + H_2SO_4 \to ZnSO_4 + H_2\\ n_{Zn} = n_{ZnSO_4} = n_{H_2} = \dfrac{6,72}{22,4} = 0,3(mol)\\ m_{Zn} = 0,3.65 = 19,5(gam)\\ m_{ZnSO_4} = 0,3.161 = 48,3(gam)\)
a) Zn+H2SO4→ZnSO4+H2
b) Ta có : VH2= 6,72(l) → nH2= nZn=nZnSO4=\(\dfrac{6,72}{22,4}\)=0,3(mol)
mZn=0,3.65=19,5(gam)
mZnSO4=0,3.161=48,3(gam)
a) PTHH: Fe + H2SO4 ===> FeSO4 + H2
b) Ta có: nFe =
Theo PTHH, nH2SO4 = nFe = 0,25 (mol)
=> mH2SO4 = 0,25 x 98 = 24,5 (gam)
c) Theo PTHH, nH2 = nFe = 0,25 (mol)
=> VH2(đktc) = 0,25 x 22,4 = 5,6 (l)
d) Theo PTHH, nFeSO4 = nFe = 0,25 (mol)
=> mFeSO4(tạo thành) = 0,25 x 152 = 38 (gam)
\(PTHH:Fe+H_2SO_4->FeSO_4+H_2\)
0,4--->0,4-------->0,4-------->0,4 (mol)
\(n_{Fe}=\dfrac{m}{M}=\dfrac{22,4}{56}=0,4\left(mol\right)\)
\(V_{H_2\left(dktc\right)}=n\cdot22,4=0,4\cdot22,4=8,96\left(l\right)\)
\(m_{H_2SO_4}=n\cdot M=0,4\cdot\left(2+32+16\cdot4\right)=39,2\left(g\right)\)
\(m_{FeSO_4}=n\cdot M=0,4\cdot\left(56+32+16\cdot4\right)=60,8\left(g\right)\)
\(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\\
pthh:Zn+2HCl\rightarrow ZnCl_2+H_2\)
0,1 0,2 0,1 0,1
\(m_{HCl}=0,2.36,5=7,3\left(g\right)\\
V_{H_2}=0,1.22,4=2,24l\\
m_{\text{dd}}=6,5+200-\left(0,1.2\right)=206,3g\)
bài 2 :
\(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\\
pthh:Mg+2HCl\rightarrow MgCl_2+H_2\)
0,2 0,4 0,2 0,2
\(m_{HCl}=0,4.36,5=14,6g\\
V_{H_2}=0,2.22,4=4,48l\\
m\text{dd}=4,8+200-0,4=204,4g\\
C\%=\dfrac{0,2.136}{204,4}.100\%=13,3\%\)
a, Cu không tác dụng với dd HCl.
PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
b, Ta có: \(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
Theo PT: \(n_{Zn}=n_{H_2}=0,2\left(mol\right)\)
\(\Rightarrow m_{Zn}=0,2.65=13\left(g\right)\)
\(\Rightarrow m_{Cu}=19,4-13=6,4\left(g\right)\)
c, Ta có: \(\left\{{}\begin{matrix}\%m_{Zn}=\dfrac{13}{19,4}.100\%\approx67,01\%\\\%m_{Cu}\approx32,99\%\end{matrix}\right.\)
Bạn tham khảo nhé!
Sửa lại câu c .
\(n_{H_2SO_4}=\dfrac{49.40}{100}:98=0,2\left(mol\right)\)
\(PTHH:\)
\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\uparrow\)
trc p/u : 0,3 0,2
p/u : 0,2 0,2 0,2 0,2
sau : 0,1 0 0,2 0,2
-> Fe dư
\(m_{ddFeSO_4}=0,3.56+49-0,4=65,4\left(g\right)\) ( ĐLBTKL )
\(m_{FeSO_4}=0,2.152=30,4\left(g\right)\)
\(C\%=\dfrac{30,4}{65,4}.100\%\approx46,48\%\)
PTHH :
\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\uparrow\)
0,3 0,3 0,3 0,3
\(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
\(a,m_{Fe}=0,3.56=16,8\left(g\right)\)
\(b,C_M=\dfrac{n}{V}=\dfrac{0,3}{0,2}=1,5M\)
\(c,n_{H_2SO_4}=\dfrac{\dfrac{49.40}{100}}{98}=0,2\left(mol\right)\)
\(\rightarrow n_{FeSO_4}=n_{H_2SO_4}=0,2\left(mol\right)\)
\(m_{FeSO_4}=0,2.152=30,4\left(g\right)\)
\(m_{ddFeSO_4}=49+\left(0,2.56\right)-0,2.2=59,8\left(g\right)\)( định luật bảo toàn khối lượng )
\(C\%=\dfrac{30,4}{59,8}.100\%\approx50,84\%\)
\(n_{H_2\left(dktc\right)}=\dfrac{V}{22,4}=\dfrac{1,792}{22,4}=0,08\left(mol\right)\)
đặt \(\left\{{}\begin{matrix}n_{Zn}=a\left(mol\right)\\n_{Al}=b\left(mol\right)\end{matrix}\right.\)
\(PTHH:Zn+H_2SO_4->ZnSO_4+H_2\)
tỉ lệ 1 : 1 : 1 : 1
n(mol) a---->a------------>a---------->a (1)
\(PTHH:2Al+3H_2SO_4->Al_2\left(SO_4\right)_3+3H_2\)
tỉ lệ 2 : 3 ; 1 : 3
n(mol) b-------->3/2b----->1/2b------------>3/2b (2)
Từ (1) và (2) ta có
\(\left\{{}\begin{matrix}65a+27b=3,79\\a+\dfrac{3}{2}b=0,08\end{matrix}\right.=>\left\{{}\begin{matrix}a=0,05\left(mol\right)\\b=0,02\left(mol\right)\end{matrix}\right.\)
\(=>\left\{{}\begin{matrix}m_{Zn}=n\cdot M=0,05\cdot65=3,25\left(g\right)\\m_{Al}=n\cdot M=0,02\cdot27=0,54\left(g\right)\end{matrix}\right.\)
\(=>\left\{{}\begin{matrix}\%m_{Zn}=\dfrac{3,25\cdot100\%}{3,79}\approx85,75\%\\\%m_{Al}=100\%-85,75\approx14,25\%\end{matrix}\right.\)
với (1) thì
\(n_{H_2SO_4\left(1\right)}=a=0,05\left(mol\right)\)
với (2) thì
\(n_{H_2SO_4\left(2\right)}=\dfrac{3}{2}b=\dfrac{3}{2}\cdot0,02=0,03\left(mol\right)\)
\(=>m_{H_2SO_4}=\left(0,05+0,03\right)\cdot98=7,84\left(g\right)\)
PTHH: \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\uparrow\)
Ta có: \(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)=n_{H_2SO_4}=n_{H_2}\)
\(\Rightarrow\left\{{}\begin{matrix}m_{H_2SO_4}=0,1\cdot98=9,8\left(g\right)\\V_{H_2}=0,1\cdot22,4=2,24\left(l\right)\end{matrix}\right.\)
\(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)
Pt : \(Fe+H_2SO_4\rightarrow FeSO_4+H_2|\)
1 1 1 1
0,1 0,1 0,1
a) \(n_{H2SO4}=\dfrac{0,1.1}{1}=0,1\left(mol\right)\)
⇒ \(m_{H2SO4}=0,1.98=9,8\left(g\right)\)
b) \(n_{H2}=\dfrac{0,1.1}{1}=0,1\left(mol\right)\)
\(V_{H2\left(dktc\right)}=0,1.22,4=2,24\left(l\right)\)
Chúc bạn học tốt
Bài 1:
nFe = \(\dfrac{1,12}{56}=0,02\) mol
Pt: 2Fe + 6H2SO4 (đ,n) --> Fe2(SO4)3 + 3SO2 + 6H2O
0,02 mol---------------------> 0,01 mol--> 0,03 mol
mFe2(SO4)3 = 0,01 . 400 = 4 (g)
VSO2 = 0,03 . 22,4 = 0,672 (lít)
Bài 2:
nNO = \(\dfrac{4,48}{22,4}=0,2\) mol
Pt: 3Zn + 8HNO3 (loãng) --> 3Zn(NO3)2 + 2NO + 4H2O
....0,3 mol<--------------------------------------0,2 mol
mZn pứ = 0,3 . 65 = 19,5 (g)
đặc nóng nha bn