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mddH2SO4=4,9%.200=9,8(g)
-> nH2SO4=9,8/98=0,1(mol)
PTHH: Zn + H2SO4 ->ZnSO4 + H2
nH2=nH2SO4=0,1(mol)
=> V(H2,đktc)=0,1.22,4=2,24(l)
\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)
\(PTHH:Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)
(mol)_____0,2____0,2______0,2____0,2__
\(a.V_{H_2}=22,4.0,2=4,48\left(l\right)\)
\(b.m_{ddH_2SO_4}=\dfrac{0,2.98.100}{24,5}=80\left(g\right)\)
\(c.m_{ddspu}=13+80-0,2.2=92,6\left(g\right)\\ \Rightarrow C\%_{ddspu}=\dfrac{0,2.136}{92,6}.100=29,4\left(\%\right)\)
\(nZn=\dfrac{13}{65}=0,2\left(mol\right)\)
\(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)
1 1 1 1 (mol)
0,2 0,2 0,2 0,2 (mol)
\(VH_2=0,2.22,4=4,48\left(l\right)\)
\(Fe_2O_3+3H_2\rightarrow2Fe+3H_2O\)
1 3 2 3 (mol)
0,2 2/15 (mol)
\(mFe=\dfrac{2}{15}.56=7,47\left(g\right)\)
a. \(n_{Zn}=\dfrac{6.5}{65}=0,1\left(mol\right)\)
PTHH : Zn + 2HCl -> ZnCl2 + H2
0,1 0,2 0,1
b. \(V_{H_2}=0,1.22,4=2,24\left(l\right)\)
c. \(m_{HCl}=0,2.36,5=7,3\left(g\right)\)
\(n_{Zn}=\dfrac{6,5}{65}=0,1mol\)
\(Zn+2HCl\rightarrow ZnCl_2+H_2\)
0,1 0,2 0,1
\(V_{H_2}=0,1\cdot22,4=2,24l\)
\(m_{HCl}=0,2\cdot36,5=7,3g\)
Câu 1 :
\(n_{H_2}=\dfrac{2.24}{22.4}=0.1\left(mol\right)\)
\(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)
\(0.1.....................................0.1\)
\(m_{Zn}=0.1\cdot65=6.5\left(g\right)\)
Câu 2 :
\(n_{Fe}=\dfrac{11.2}{56}=0.2\left(mol\right)\)
\(n_{HCl}=\dfrac{10.95}{36.5}=0.3\left(mol\right)\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
\(1..............2\)
\(0.2...........0.3\)
\(LTL:\dfrac{0.2}{1}>\dfrac{0.3}{2}\Rightarrow Fedư\)
\(m_{Fe\left(dư\right)}=\left(0.2-0.15\right)\cdot56=2.8\left(g\right)\)
\(V_{H_2}=0.15\cdot22.4=3.36\left(l\right)\)
\(n_{Zn}=\dfrac{9,75}{65}=0,15\left(mol\right)\\ pthh:Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)
0,15 0,15 0,15
\(V_{H_2}=0,15.22,4=3,36\left(l\right)\\
C_M=\dfrac{0,15}{0,1}=1,5M\)
1.
a, \(n_{Al}=\dfrac{8,1}{27}=0,3\left(mol\right)\)
PTHH: 2Al + 3H2SO4 → Al2(SO4)3 + 3H2
Mol: 0,3 0,15 0,45
b, \(V_{H_2}=0,45.22,4=10,08\left(l\right)\)
c, Al2(SO4)3 : nhôm sunfat
\(m_{Al_2\left(SO_4\right)_3}=0,15.342=51,3\left(g\right)\)
3.
a, \(n_{Cu}=\dfrac{38,4}{64}=0,6\left(mol\right)\)
PTHH: 2Cu + O2 ---to→ 2CuO
Mol: 0,6 0,3
CuO: đồng(ll) oxit
b, \(V_{O_2}=0,3.22,4=6,72\left(l\right)\)
c,
PTHH: 2KMnO4 ---to→ K2MnO4 + MnO2 + O2
Mol: 0,6 0,3
\(m_{KMnO_4}=0,6.158=47,4\left(g\right)\)
PTHH: \(Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)
Ta có: \(n_{Zn}=\dfrac{0,65}{65}=0,01\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}n_{HCl}=0,02\left(mol\right)\\n_{ZnCl_2}=0,01\left(mol\right)=n_{H_2}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}C_{M_{HCl}}=\dfrac{0,02}{0,05}=0,4\left(M\right)\\m_{ZnCl_2}=0,01\cdot136=1,36\left(g\right)\\V_{H_2}=0,01\cdot22,4=0,224\left(l\right)\end{matrix}\right.\)
a. Zn + 2HCl → ZnCl2 + H2
b. nZn = n\(_{ZnCl_2}\) =\(\dfrac{13}{65}=0,2\left(mol\right)\) => m\(_{ZnCl_2}\)= 0,2.136 = 27,2(g)
c. n\(_{H_2}\)= nZn = 0,2 (mol) => V\(_{H_2}\)=0,2.22,4 = 4,48 (lít)
Câu 1 :
$S = \dfrac{15}{100}.100 = 15(gam)$
Đáp án A
Câu 2 :
Đáp án A ($SO_2$)
Câu 3 :
$Zn + H_2SO_4 \to ZnSO_4 + H_2$
$n_{H_2} = n_{H_2SO_4} = \dfrac{200.4,9\%}{98} = 0,1(mol)$
$V_{H_2} = 0,1.22,4 = 2,24(lít)$
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