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a) PTHH: \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\uparrow\)
b+c)
Ta có: \(n_{Zn}=\dfrac{19,5}{65}=0,3\left(mol\right)=n_{H_2SO_4}=n_{H_2}\)
\(\Rightarrow\left\{{}\begin{matrix}V_{H_2}=0,3\cdot22,4=6,72\left(l\right)\\m_{ddH_2SO_4}=\dfrac{0,3\cdot98}{30\%}=98\left(g\right)\end{matrix}\right.\)
d) PTHH: \(ZnSO_4+BaCl_2\rightarrow ZnCl_2+BaSO_4\downarrow\)
Ta có: \(n_{BaCl_2}=\dfrac{260\cdot20\%}{208}=0,25\left(mol\right)\)
Xét tỉ lệ: \(\dfrac{0,3}{1}>\dfrac{0,25}{1}\) \(\Rightarrow\) ZnSO4 còn dư, BaCl2 phản ứng hết
\(\Rightarrow\left\{{}\begin{matrix}n_{ZnCl_2}=0,25mol=n_{BaSO_4}\\n_{ZnSO_4\left(dư\right)}=0,05mol\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{ZnCl_2}=0,25\cdot136=34\left(g\right)\\m_{BaSO_4}=0,25\cdot233=58,25\left(g\right)\\m_{ZnSO_4\left(dư\right)}=0,05\cdot161=8,05\left(g\right)\end{matrix}\right.\)
Mặt khác: \(m_{H_2}=0,3\cdot2=0,6\left(g\right)\)
\(\Rightarrow m_{dd}=m_{Zn}+m_{ddH_2SO_4}-m_{H_2}+m_{ddBaCl_2}-m_{BaSO_4}=318,65\left(g\right)\)
\(\Rightarrow\left\{{}\begin{matrix}C\%_{ZnCl_2}=\dfrac{34}{318,65}\cdot100\%\approx10,67\%\\C\%_{ZnSO_4\left(dư\right)}=\dfrac{8,05}{318,65}\cdot100\%\approx2,53\%\end{matrix}\right.\)
\(a,n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
\(PTHH:Zn+H_2SO_4\rightarrow ZnSO_4+H_2\\ ..........0,15.......0,15.......0,15.......0,15\left(mol\right)\)
\(m_{Zn}=65\cdot0,15=9,75\left(g\right)\)
\(b,m_{H_2SO_4}=98\cdot0,15=14,7\left(mol\right)\\ c,m_{dd_{H_2SO_4}}=\dfrac{14,7\cdot100}{20}=\dfrac{147}{2}\left(g\right)\\ d,C\%_{dd_{ZnSO_4}}=\dfrac{0,15\cdot161}{\dfrac{147}{2}}\cdot100\approx32,86\%\)
a)
$n_{Al} = 0,3(mol)$
$2Al + 3H_2SO_4 \to Al_2(SO_4)_3 + 3H_2$
Theo PTHH :
$n_{H_2SO_4} = \dfrac{3}{2}n_{Al} = 0,45(mol)$
$m_{dd\ H_2SO_4} = \dfrac{0,45.98}{12,25\%} = 360(gam)$
b)
$n_{H_2} = n_{H_2SO_4} = 0,45(mol)$
$V_{H_2} = 0,45.22,4 = 10,08(lít)$
c)
$n_{Al_2(SO_4)_3} = 0,15(mol)$
$m_{dd\ sau\ pư} = 8,1 + 360 - 0,45.2 = 367,2(gam)$
$C\%_{Al_2(SO_4)_3} = \dfrac{0,15.342}{367,2}.100\% = 14\%$
a) PTHH: Zn + H2SO4 -> ZnSO4 + H2
nH2= 0,15(mol)
=> nZn=nH2SO4=nZnSO4=nH2=0,15(mol)
b) mZn=0,15.65=9,75(g)
c) CMddH2SO4= 0,15/ 0,05=3(M)
d) mZnSO4= 161. 0,15=24,15(g)
Chỉ có Zn phản ứng thôi. Cu không phản ứng, không tan.---->Chất rắn không tan là Cu
Zn+ H2SO4 ---> ZnSO4+ H2↑
0.1 0.1
nH2= 2.24: 22.4=0.1 mol
mZn= 0.1x65=6.5 g
mCu=10.5-6,5=4 g
%Zn=6.5:10.5x100%=61.9%
%Cu=4:10.5x100%=38.1%
a, \(Na_2SO_3+H_2SO_4\rightarrow Na_2SO_4+SO_2+H_2O\)
\(K_2SO_3+H_2SO_4\rightarrow K_2SO_4+SO_2+H_2O\)
Ta có: \(n_{SO_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
Theo PT: \(n_{H_2SO_4}=n_{SO_2}=0,3\left(mol\right)\)
\(\Rightarrow m_{ddH_2SO_4}=\dfrac{0,3.98}{20\%}=147\left(g\right)\)
b, Ta có: 126nNa2SO3 + 158nK2SO3 = 44,2 (1)
Theo PT: \(n_{SO_2}=n_{Na_2SO_3}+n_{K_2SO_3}=0,3\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}n_{Na_2SO_3}=0,1\left(mol\right)\\n_{K_2SO_3}=0,2\left(mol\right)\end{matrix}\right.\)
Có: m dd sau pư = 44,2 + 147 - 0,3.64 = 172 (g)
\(\Rightarrow\left\{{}\begin{matrix}C\%_{Na_2SO_3}=\dfrac{0,1.126}{172}.100\%\approx7,33\%\\C\%_{K_2SO_3}=\dfrac{0,2.158}{172}.100\%\approx18,37\%\end{matrix}\right.\)
c, \(n_{Ba\left(OH\right)_2}=0,5.1=0,5\left(mol\right)\)
\(\Rightarrow\dfrac{n_{SO_2}}{n_{Ba\left(OH\right)_2}}=0,6< 1\) → Pư tạo BaSO3.
PT: \(SO_2+Ba\left(OH\right)_2\rightarrow BaSO_3+H_2O\)
\(n_{BaSO_3}=n_{SO_2}=0,3\left(mol\right)\Rightarrow m_{BaSO_3}=0,3.217=65,1\left(g\right)\)
a)\(PTHH:Fe+H_2SO_4\rightarrow FeSO_4+H_2\)
\(Mg+H_2SO_4\rightarrow MgSO_4+H_2\)
\(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)
\(n_{H_2}=\frac{2,688}{22,4}=0,12\left(mol\right)\)
\(n_{SO^{ }_4}=\frac{1}{2}n_{H_2}=0,06\left(mol\right)\)
\(m_{muối}=m_{KL}+m_{SO_4}=6,44+0,06.96=12,2\left(g\right)\)
b) \(n_{H_2SO_4}=n_{H_2}=0,12\left(mol\right)\)
\(m_{ddH_2SO_4}=\frac{0,12.98}{9,8\%}=120\left(g\right)\)
Ta có: \(n_{Na_2SO_3}=\dfrac{12,6}{126}=0,1\left(mol\right)\)
a. PTHH: Na2SO3 + 2HCl ---> 2NaCl + SO2 + H2O
Theo PT: \(n_{SO_2}=n_{Na_2CO_3}=0,1\left(mol\right)\)
=> \(V_{SO_2}=0,1.22,4=2,24\left(lít\right)\)
b. Theo PT: \(n_{HCl}=2.n_{SO_2}=2.0,1=0,2\left(mol\right)\)
=> \(m_{HCl}=0,2.36,5=7,3\left(g\right)\)
=> \(C_{\%_{HCl}}=\dfrac{7,3}{150}.100\%=4,87\%\)
c. Ta có: \(m_{dd_{NaCl}}=n_{Na_2SO_{3_{PỨ}}}=50\left(g\right)\)
Theo PT: \(n_{NaCl}=n_{HCl}=0,2\left(mol\right)\)
=> \(m_{NaCl}=0,2.58,5=11,7\left(g\right)\)
=> \(C_{\%_{NaCl}}=\dfrac{11,7}{50}.100\%=23,4\%\)
\(a.n_{Zn}=\dfrac{32,5}{65}=0,5\left(mol\right)\\ PTHH:Zn+H_2SO_4\rightarrow ZnSO_4+H_2\\ 0,5..........0,5...........0,5........0,5\left(mol\right)\\ b.V_{H_2\left(đktc\right)}=0,5.22,4=11,2\left(l\right)\\ c.m_{ddH_2SO_4}=\dfrac{0,5.98.100}{9}=\dfrac{4900}{9}\left(g\right)\\ d.C\%_{ddZnSO_4}=\dfrac{0,5.161}{\dfrac{4900}{9}}.100\approx14,786\%\)