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a, \(MgCO_3+2HCl\rightarrow MgCl_2+CO_2+H_2O\)
Ta có: \(n_{CO_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
Theo PT: \(n_{MgCO_3}=n_{CO_2}=0,2\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{MgCO_3}=\dfrac{0,2.84}{64,8}.100\%\approx25,93\%\\\%m_{MgSO_4}\approx74,07\%\end{matrix}\right.\)
b, - Dung dịch C gồm: MgCl2, MgSO4 và HCl dư.
Theo PT: \(\left\{{}\begin{matrix}n_{MgCl_2}=n_{CO_2}=0,2\left(mol\right)\\n_{HCl\left(pư\right)}=2n_{CO_2}=0,4\left(mol\right)\end{matrix}\right.\)
Ta có: \(m_{HCl}=100.18,25\%=18,25\left(g\right)\Rightarrow n_{HCl}=\dfrac{18,25}{36,5}=0,5\left(mol\right)\)
\(\Rightarrow n_{HCl\left(dư\right)}=0,5-0,4=0,1\left(mol\right)\)
\(m_{MgSO_4}=64,8-0,2.84=48\left(g\right)\Rightarrow n_{MgSO_4}=\dfrac{48}{120}=0,4\left(mol\right)\)
Có: m dd sau pư = 64,8 + 100 - 0,2.44 = 156 (g)
\(\Rightarrow\left\{{}\begin{matrix}C\%_{MgCl_2}=\dfrac{0,2.95}{156}.100\%\approx12,18\%\\C\%_{HCl\left(dư\right)}=\dfrac{0,1.36,5}{156}.100\%\approx2,34\%\\C\%_{MgSO_4}=\dfrac{48}{156}.100\%\approx30,77\%\end{matrix}\right.\)
c, PT: \(MgCl_2+2NaOH\rightarrow2NaCl+Mg\left(OH\right)_{2\downarrow}\)
\(HCl+NaOH\rightarrow NaCl+H_2O\)
\(MgSO_4+2NaOH\rightarrow Na_2SO_4+Mg\left(OH\right)_{2\downarrow}\)
\(Mg\left(OH\right)_2\underrightarrow{t^o}MgO+H_2O\)
Theo PT: \(n_{MgO}=n_{Mg\left(OH\right)_2}=n_{MgCl_2}+n_{MgSO_4}=0,6\left(mol\right)\)
\(\Rightarrow m_{cr}=m_{MgO}=0,6.40=24\left(g\right)\)
PTHH: \(K_2CO_3+2HCl\rightarrow2KCl+CO_2\uparrow+H_2O\)
Ta có: \(n_{HCl}=\dfrac{200\cdot7,3\%}{36,5}=0,4\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}n_{K_2CO_3}=n_{CO_2}=0,2\left(mol\right)\\n_{KCl}=0,4\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{ddK_2CO_3}=\dfrac{0,2\cdot138}{13,8\%}=200\left(g\right)\\V_{CO_2}=0,2\cdot22,4=4,48\left(l\right)\\m_{KCl}=0,4\cdot74,5=29,8\left(g\right)\end{matrix}\right.\)
\(a) n_{CH_3COOH} = \dfrac{200.12\%}{60} = 0,4(mol)\\ 2CH_3COOH + CaCO_3 \to (CH_3COO)_2Ca + CO_2 + H_2O\\ n_{CaCO_3} = n_{CO_2} = \dfrac{1}{2}n_{CH_3COOH} = 0,2(mol)\\ \Rightarrow a = \dfrac{0,2.100}{100\%-20\%} =25(gam)\\ V_B = 0,2.22,4 = 4,48(lít)\\ b) m_{dd\ sau\ pư} = m_{CaCO_3} + m_{dd\ CH_3COOH} - m_{CO_2} = 0,2.100 + 200 - 0,2.2 = 219,6(gam)\\ C\%_{(CH_3COO)_2Ca} = \dfrac{0,2.158}{219,6}.100\% = 36\%\)
\(Đặt:n_{Na_2CO_3}=a\left(mol\right);n_{K_2CO_3}=b\left(mol\right)\\ Na_2CO_3+2HCl\rightarrow2NaCl+CO_2+H_2O\\ K_2CO_3+2HCl\rightarrow2KCl+CO_2+H_2O\\ CO_2+Ca\left(OH\right)_2\rightarrow CaCO_3\downarrow+H_2O\\ n_{CaCO_3}=n_{CO_2}=0,3\left(mol\right)\\ \Rightarrow\left\{{}\begin{matrix}106a+138b=38,2\\a+b=0,3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,1\\b=0,2\end{matrix}\right.\\ a.C\%_{ddHCl}=\dfrac{0,6.36,5}{200}.100=10,95\%\\ b.m_{ddB}=38,2+200-0,3.44=225\left(g\right)\\ C\%_{ddKCl}=\dfrac{74,5.2.0,2}{225}.100\approx13,244\%\\ C\%_{ddNaCl}=\dfrac{58,5.2.0,1}{225}.100=5,2\%\)
- Cả 2 chất trong hhA đều tác dụng được với dd HCl dư. Nhưng chỉ có Zn tác dụng với dd HCl dư mới sinh ra khí H2
\(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\\ PTHH:\left(1\right)Zn+2HCl\rightarrow ZnCl_2+H_2\\ \left(2\right)ZnO+2HCl\rightarrow ZnCl_2+H_2O\\ TheoPTHH\left(1\right):n_{Zn}=n_{ZnCl_2\left(1\right)}=n_{H_2}=0,2\left(mol\right)\\ m_{ZnO}=m_{hhA}-m_{Zn}=21,1-65.0,2=8,1\left(g\right)\\ n_{ZnO}=\dfrac{8,1}{81}=0,1\left(mol\right)\\ n_{ZnCl_2\left(2\right)}=n_{ZnO}=0,1\left(mol\right)\\ n_{ZnCl_2\left(tổng\right)}=0,2+0,1=0,3\left(mol\right)\\ m_{ddB}=m_{hhA}+m_{ddHCl}-m_{H_2}=21,1+200-0,2.2=220,7\left(g\right)\\ C\%_{ddZnCl_2}=\dfrac{136.0,3}{220,7}.100\%\approx18,487\%\)
\(n_{H_2}=\dfrac{4,48}{22,4}=0,2mol\\ Zn+2HCl\rightarrow ZnCl_2+H_2\left(1\right)\\ n_{Zn}=n_{H_2}=n_{ZnCl_2\left(1\right)}=0,2mol\\ n_{ZnO}=\dfrac{21,1-0,2.65}{81}=0,1mol\\ ZnO+2HCl\rightarrow ZnCl_2+H_2O\left(2\right)\\ n_{ZnCl_2\left(2\right)}=n_{ZnO}=0,1mol\\ C_{\%B}=C_{\%ZnCl_2}=\dfrac{\left(0,2+0,1\right).136}{21,1+200-0,2.2}\cdot100\%=18,49\%\)
a) $Fe + 2HCl \to FeCl_2 + H_2$
b) $n_{Fe} = \dfrac{11,2}{56} = 0,2(mol)$
$n_{HCl} = \dfrac{200.7,3\%}{36,5} = 0,4(mol)$
$Fe + 2HCl \to FeCl_2 + H_2$
Ta thấy : $n_{Fe} : 1 = n_{HCl} : 2$ nên phản ứng vừa đủ
$n_{H_2} = n_{Fe} = 0,2(mol)$
$V_{H_2} = 0,2.22,4 = 4,48(lít)$
c) $m_{dd\ sau\ pư} = 11,2 + 200 - 0,2.2 = 210,8(gam)$
$C\%_{FeCl_2} = \dfrac{0,2.127}{210,8}.100\% = 12,05\%$
\(n_{Na_2CO_3}=\dfrac{200.0,159}{106}=0,3mol\\ n_{BaCl_2}=\dfrac{200.0,208}{208}=0,2mol\\ a.Na_2CO_3+BaCl_2->2NaCl+BaCO_3\\ n_{Na_2CO_3}:1>n_{BaCl_2}:1\\ m_B=197.0,2=39,4g\\ Na_2CO_3+2HCl->2NaCl+H_2O+CO_2\\ V=\dfrac{2.0,1}{1}=0,2\left(L\right)=200\left(mL\right)\\ b.m_A=200+200-39,4=360,6g\\ C\%_{Na_2CO_3du}=\dfrac{106.0,1}{360,6}.100\%=2,94\%\\ C\%_{NaCl}=\dfrac{58,5.0,4}{360,6}.100\%=6,49\%\)
a)
\(Na_2CO_3+BaCl_2\rightarrow BaCO_3+2NaCl\)
0,2 <---------- 0,2 ------> 0,2 -----> 0,4
\(n_{Na_2CO_3}=\dfrac{200.15,9\%}{100\%}:106=0,3\left(mol\right)\)
\(n_{BaCl_2}=\dfrac{200.20,8\%}{100\%}:208=0,2\left(mol\right)\)
Do \(\dfrac{0,3}{1}>\dfrac{0,2}{1}\) nên \(Na_2CO_3\) dư sau phản ứng.
Dung dịch A: \(n_{Na_2CO_3}=0,3-0,2=0,1\left(mol\right);n_{NaCl}:0,4\left(mol\right)\)
Kết tủa B: \(BaCO_3\)
\(m_B=m_{BaCO_3}=0,2.197=39,4\left(g\right)\)
Dung dịch A td với HCl:
\(Na_2CO_3+2HCl\rightarrow2NaCl+H_2O+CO_2\)
0,1 ---------> 0,2
\(V=V_{HCl}=\dfrac{0,2}{1}=0,2\left(l\right)\)
b)
\(m_{dd}=m_{dd.Na_2CO_3}+m_{dd.BaCl_2}-m_{BaCO_3}=200+200-39,4=360,6\left(g\right)\)
\(C\%_{Na_2CO_3}=\dfrac{0,1.106.100\%}{360,6}=2,94\%\)
\(C\%_{NaCl}=\dfrac{0,4.58,5.100\%}{360,6}=6,49\%\)
a) Ta có: \(\left\{{}\begin{matrix}n_{K_2CO_3}=\dfrac{200.13,8\%}{138}=0,2\left(mol\right)\\n_{HCl}=\dfrac{100.18,25\%}{36,5}=0,5\left(mol\right)\end{matrix}\right.\)
PTHH: \(K_2CO_3+2HCl\rightarrow2KCl+CO_2+H_2O\)
ban đầu 0,2 0,5
phản ứng 0,2-------->0,4
sau phản ứng 0 0,1 0,4 0,2
`=> V_{B} = V_{CO_2} = 0,2.22,4 = 4,48 (l)`
b) `m_{dd} = 200 + 100 - 0,2.44 = 291,2 (g)`
`=>` \(\left\{{}\begin{matrix}C\%_{KCl}=\dfrac{0,4.74,5}{291,2}.100\%=10,23\%\\C\%_{HCl.dư}=\dfrac{0,1.36,5}{291,2}.100\%=1,25\%\end{matrix}\right.\)
Cho mk hỏi cái chỗ mà mdd á thì 44 ở đâu á