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Câu 2:
a, \(n_{Fe_2O_3}=\dfrac{6,4}{160}=0,04\left(mol\right)\)
\(n_{H_2SO_4}=0,5.1=0,5\left(mol\right)\)
PT: \(Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O\)
Xét tỉ lệ: \(\dfrac{0,04}{1}< \dfrac{0,5}{3}\), ta được H2SO4 dư.
Vậy: Fe2O3 tan hết.
b, Theo PT: \(\left\{{}\begin{matrix}n_{Fe_2\left(SO_4\right)_3}=n_{Fe_2O_3}=0,04\left(mol\right)\\n_{H_2SO_4\left(pư\right)}=3n_{Fe_2O_3}=0,12\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow n_{H_2SO_4\left(dư\right)}=0,5-0,12=0,38\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}C_{M_{Fe_2\left(SO_4\right)_3}}=\dfrac{0,04}{0,5}=0,08\left(M\right)\\C_{M_{H_2SO_4\left(dư\right)}}=\dfrac{0,38}{0,5}=0,76\left(M\right)\end{matrix}\right.\)
Câu 3:
a, \(n_{Ba\left(OH\right)_2}=0,2.1=0,2\left(mol\right)\)
\(n_{H_2SO_4}=0,3.0,72=0,216\left(mol\right)\)
PT: \(Ba\left(OH\right)_2+H_2SO_4\rightarrow BaSO_{4\downarrow}+2H_2O\)
Xét tỉ lệ: \(\dfrac{0,2}{1}< \dfrac{0,216}{1}\), ta được H2SO4 dư.
Theo PT: \(n_{BaSO_4}=n_{Ba\left(OH\right)_2}=0,2\left(mol\right)\Rightarrow m_{BaSO_4}=0,2.233=46,6\left(g\right)\)
b, Theo PT: \(n_{H_2SO_4\left(pư\right)}=n_{Ba\left(OH\right)_2}=0,2\left(mol\right)\Rightarrow n_{H_2SO_4\left(dư\right)}=0,216-0,2=0,016\left(mol\right)\)
\(\Rightarrow C_{M_{H_2SO_4\left(dư\right)}}=\dfrac{0,016}{0,2+0,3}=0,032\left(M\right)\)
c, - Nhúng quỳ tím vào dd thấy quỳ hóa đỏ do H2SO4 dư.
Bài 1 :
\(a) CaCO_3 \xrightarrow{t^o} CaO + CO_2\\ 2Fe(OH)_3 \xrightarrow{t^o} Fe_2O_3 + 3H_2O\\ b) CaCO_3 + 2HCl \to CaCl_2 + CO_2 + H_2O\\ 2Fe(OH)_3 + 6HCl \to 2FeCl_3 + 6H_2O\\ NaOH + HCl \to NaCl + H_2O\\ c) 2AgNO_3 + 2NaOH \to 2NaNO_3 + Ag_2O + H_2O\\ NaCl + AgNO_3 \to AgCl + NaNO_3\)
Ta có: \(m_{ddH_2SO_4\left(60\%\right)}=700.1,503=1052,1\left(g\right)\Rightarrow m_{H_2SO_4}=1052,1.60\%=631,26\left(g\right)\)
\(m_{ddH_2SO_4\left(20\%\right)}=500.1,1476=573,8\left(g\right)\Rightarrow m_{H_2SO_4}=573,8.20\%=114,76\left(g\right)\)
ΣmH2SO4 = 631,26 + 114,76 = 746,02 (g)
\(n_{H_2}=\dfrac{1,792}{22,4}=0,08\left(mol\right)\)
PT: \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)
Theo PT: \(n_{H_2SO_4}=n_{H_2}=0,08\left(mol\right)\)
\(\Rightarrow m_{H_2SO_4}=0,08.98=7,84\left(g\right)\)
\(\Rightarrow\dfrac{746,02}{V}=\dfrac{7,84}{0,2}\Rightarrow V\approx19,03\left(l\right)\)
\(n_{Ba}=\dfrac{41,1}{137}=0,3\left(mol\right)\\ PTHH:Ba+H_2SO_4\rightarrow BaSO_4+H_2\\ n_{H_2SO_4}=0,2.1=0,2\left(mol\right)\\ Vì:\dfrac{0,2}{1}< \dfrac{0,3}{1}\)
=> Ba dư
\(n_{H_2}=n_{H_2SO_4}=0,2\left(mol\right)\\ V_{H_2\left(đktc\right)}=0,2.24,79=4,958\left(l\right)\)
\(n_{H_2SO_4}=0.1\cdot2=0.2\left(mol\right)\)
\(m_{dd_{H_2SO_4}}=100\cdot1.2=120\left(g\right)\)
\(n_{BaCl_2}=0.1\cdot1=0.1\left(mol\right)\)
\(m_{dd_{BaCl_2}}=100\cdot1.32=132\left(g\right)\)
\(BaCl_2+H_2SO_4\rightarrow BaSO_4+2HCl\)
\(0.1................0.1.........0.1...............0.2\)
\(\Rightarrow H_2SO_4dư\)
\(m_{BaSO_4}=0.1\cdot233=23.3\left(g\right)\)
\(V_{dd}=0.1+0.1=0.2\left(l\right)\)
\(C_{M_{H_2SO_4\left(dư\right)}}=\dfrac{0.2-0.1}{0.2}=0.5\left(M\right)\)
\(C_{M_{HCl}}=\dfrac{0.2}{0.2}=1\left(M\right)\)
\(m_{\text{dung dịch sau phản ứng}}=120+132-23.3=228.7\left(g\right)\)
\(C\%_{H_2SO_4\left(dư\right)}=\dfrac{0.1\cdot98}{228.7}\cdot100\%=4.28\%\)
\(C\%_{HCl}=\dfrac{0.2\cdot36.5}{228.7}\cdot100\%=3.2\%\)