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PTHH:\(Na_2SO_3+CaCl_2\rightarrow2NaCl+CaSO_3\downarrow\)
Ta có: \(\left\{{}\begin{matrix}n_{Na_2SO_3}=\dfrac{265\cdot10\%}{126}=\dfrac{53}{252}\left(mol\right)\\n_{CaCl_2}=\dfrac{500\cdot6,66\%}{111}=0,3\left(mol\right)\end{matrix}\right.\)
Xét tỷ số: \(\dfrac{53}{252}< \dfrac{0,3}{1}\) \(\Rightarrow\) CaCl2 còn dư, Na2SO3 phản ứng hết
\(\Rightarrow\left\{{}\begin{matrix}n_{NaCl}=\dfrac{53}{126}\left(mol\right)\\n_{CaSO_3}=\dfrac{53}{252}\left(mol\right)\\n_{CaCl_2\left(dư\right)}=\dfrac{113}{1260}\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{NaCl}=\dfrac{53}{126}\cdot58,5\approx24,61\left(g\right)\\m_{CaSO_3}=\dfrac{53}{252}\cdot120\approx25,24\left(g\right)\\m_{CaCl_2\left(dư\right)}=\dfrac{113}{1260}\cdot111\approx9,95\left(g\right)\end{matrix}\right.\)
Mặt khác: \(m_{dd}=m_{ddNa_2SO_3}+m_{ddCaCl_2}-m_{CaSO_3}=739,76\left(g\right)\)
\(\Rightarrow\left\{{}\begin{matrix}C\%_{NaCl}=\dfrac{24,61}{739,76}\cdot100\%\approx3,33\%\\C\%_{CaCl_2\left(dư\right)}=\dfrac{9,95}{739,76}\cdot100\%\approx1,35\%\end{matrix}\right.\)
$n_{Fe_2O_3} = 0,05(mol)$
$n_{H_2SO_4} = \dfrac{150.20\%}{98} = \dfrac{15}{49}(mol)$
$Fe_2O_3 + 3H_2SO_4 \to Fe_2(SO_4)_3 + 3H_2O$
Ta thấy :
$n_{Fe_2O_3} : 1 < n_{H_2SO_4} :3$ nên $H_2SO_4$ dư
$m_{dd\ sau\ pư} = 8 + 150 = 158(gam)$
$n_{H_2SO_4\ dư} = \dfrac{15}{49} - 0,05.3 = \dfrac{153}{980}(mol)$
$n_{Fe_2(SO_4)_3} = 0,025(mol)$
$C\%_{H_2SO_4} = \dfrac{ \dfrac{153}{980}.98}{158} .100\% = 9,7\%$
$C\%_{Fe_2(SO_4)_3} = \dfrac{0,025.400}{158}.100\% = 6,3\%$
\(n_{FeO}=\dfrac{7,2}{72}=0,1mol\\ n_{H_2SO_4}=0,4.1,5=0,6mol\\ FeO+H_2SO_4\rightarrow FeSO_4+H_2O\\ \Rightarrow\dfrac{0,1}{1}< \dfrac{0,6}{1}\Rightarrow H_2SO_4.dư\\ n_{FeO}=n_{FeSO_4}=n_{H_2SO_4,pư}=0,1mol\\ C_{M_{FeSO_4}}=\dfrac{0,1}{0,4}=0,25M\\ C_{M_{H_2SO_4}}=\dfrac{0,6-0,1}{0,4}=1,25M\)
\(m_{Na_2CO_3}\) = \(265\times10\%\) = \(26,5\left(g\right)\) \(\Rightarrow n_{Na_2CO_3}\) = 0,25
\(m_{CaCl_2}\) = \(500\times6,6\%\) = \(33\left(g\right)\) \(\Rightarrow n_{CaCl_2}\)= \(\dfrac{11}{37}\)
\(Na_2CO_3+CaCl_2=2NaCl+CaCO_3\)↓
0,25_______11/37__0,5______0,25
dd sau phản ứng gồm \(NaCl\) \(0,5mol\) và \(CaCl_2\) dư = \(\dfrac{7}{148}\)mol
m dd sau phản ứng = trước phản ứng - m↓= 265 + 500 - \(0,25\times100\) = 740 (g)
=> C% dd \(NaCl\) = \(0,5\times\dfrac{58,5}{740}\times100\%\) = 3,95%
C% dd \(CaCl_2\)= \(\dfrac{7}{148}\times\dfrac{111}{740}\times100\%\) = 0,71%
\(n_{MgO}=\dfrac{2}{40}=0,05\left(mol\right)\)
\(n_{H2SO4}=\dfrac{19,6\%.100}{100\%.98}=0,2\left(mol\right)\)
Pt : \(MgO+H_2SO_4\rightarrow MgSO_4+H_2O\)
Xét tỉ lệ : \(\dfrac{0,05}{1}< \dfrac{0,2}{1}\Rightarrow H_2SO_4dư\)
Theo pt : \(n_{MgO\left(pư\right)}=n_{MgSO4}=0,05\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}C\%_{MgSO4}=\dfrac{0,05.120}{2+100}.100\%=5,88\%\\C\%_{ddH2SO4\left(dư\right)}=\dfrac{\left(0,2-0,05\right).98}{2+100}.100\%=14,41\%\end{matrix}\right.\)
$n_{Na_2CO_3} = \dfrac{265.10\%}{106} = 0,25(mol)$
$n_{CaCl_2} = \dfrac{475,72.7\%}{111} = 0,3(mol)$
$CaCl_2 + Na_2CO_3 \to CaCO_3 + 2NaCl$
Ta thấy : $n_{CaCl_2} > n_{Na_2CO_3}$ nên $CaCl_2$ dư
$n_{CaCl_2\ dư} = 0,3 - 0,25 = 0,05(mol)$
$n_{NaCl} = 0,5(mol)$
Sau phản ứng, $m_{dd} = 265 + 475,72 - 0,25.100 = 715,72(gam)$
$C\%_{NaCl} = \dfrac{0,5.58,5}{715,72}.100\% = 4,09\%$
$C\%_{CaCl_2\ dư} = \dfrac{0,05.111}{715,72}.100\% = 0,76\%$
bạn cho mình hỏi chỗ này sao lại trừ đi 0,25.100
Sau phản ứng, mdd=265+475,72−0,25.100=715,72(gam)