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\(a,PTHH:H_2SO_4+BaCl_2\rightarrow BaSO_4\downarrow+2HCl\\ \left\{{}\begin{matrix}m_{H_2SO_4}=\dfrac{300\cdot9,8\%}{100\%}=29,4\left(g\right)\\m_{BaCl_2}=\dfrac{200\cdot26\%}{100\%}=52\left(g\right)\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}n_{H_2SO_4}=\dfrac{29,4}{98}=0,3\left(mol\right)\\n_{BaCl_2}=\dfrac{52}{208}=0,25\left(mol\right)\end{matrix}\right.\)
Vì \(\dfrac{n_{H_2SO_4}}{1}>\dfrac{n_{BaCl_2}}{1}\) nên H2SO4 dư
\(\Rightarrow n_{BaSO_4}=n_{BaCl_2}=0,25\left(mol\right)\\ \Rightarrow a=m_{BaSO_4}=0,25\cdot233=58,25\left(g\right)\\ b,n_{HCl}=n_{BaCl_2}=0,25\left(mol\right)\\ \Rightarrow m_{CT_{HCl}}=0,25\cdot36,5=9,125\left(g\right)\\ m_{dd_{HCl}}=300+200-58,25=441,75\left(g\right)\\ \Rightarrow C\%_{HCl}=\dfrac{9,125}{441,75}\cdot100\%\approx2,07\%\)
Gọi \(\left\{{}\begin{matrix}C_{M\left(A\right)}=aM\\C_{M\left(B\right)}=bM\end{matrix}\right.\)
Giả sử trộn 50ml dd A với 50ml dd B để thu được 100ml dd C
=> \(\left\{{}\begin{matrix}n_{NaOH}=0,05a\left(mol\right)\\n_{Ba\left(OH\right)_2}=0,05b\left(mol\right)\end{matrix}\right.\)
\(n_{BaSO_4}=\dfrac{9,32}{233}=0,04\left(mol\right)\)
nH2SO4 = 0,035.2 = 0,07 (mol)
PTHH: Ba(OH)2 + H2SO4 --> BaSO4 + 2H2O
0,04<----0,04<-------0,04
2NaOH + H2SO4 --> Na2SO4 + 2H2O
0,06<----0,03
=> \(\left\{{}\begin{matrix}0,05a=0,06\\0,05b=0,04\end{matrix}\right.\)
=> a = 1,2; b = 0,8
20 ml dd A chứa nNaOH = 0,02.1,2 = 0,024 (mol)
\(n_{Al_2O_3}=\dfrac{20,4}{102}=0,2\left(mol\right)\)
PTHH: 2NaOH + Al2O3 --> 2NaAlO2 + H2O
0,024-->0,012
Ba(OH)2 + Al2O3 --> Ba(AlO2)2 + H2O
0,188<---0,188
=> \(V_{dd.B}=\dfrac{0,188}{0,8}=0,235\left(l\right)=235\left(ml\right)\)
\(m_{Ba\left(OH\right)_2}=150.17,1\%=25,65\left(g\right)\Rightarrow n_{Ba\left(OH\right)_2}=\dfrac{25,65}{171}=0,15\left(mol\right)\)
\(m_{FeSO_4}=250.15,2\%=38\left(g\right)\Rightarrow n_{FeSO_4}=\dfrac{38}{152}=0,25\left(mol\right)\)
PTHH: Ba(OH)2 + FeSO4 → Fe(OH)2 + BaSO4
Mol: 0,15 0,15 0,15 0,15
Ta có: \(\dfrac{0,15}{1}< \dfrac{0,25}{1}\) ⇒ Ba(OH)2 pứ hết, FeSO4 dư
\(m=0,15.233=34,95\left(g\right)\)
\(C\%_{Fe\left(OH\right)_2}=\dfrac{0,15.90.100\%}{400}=3,375\%\)
\(C\%_{FeSO_4dư}=\dfrac{\left(0,25-0,15\right).152.100\%}{400}=3,8\%\)
Ta có: \(m_{CuSO_4}=\dfrac{10\%.160}{100\%}=16\left(g\right)\)
=> \(n_{CuSO_4}=\dfrac{16}{160}=0,1\left(mol\right)\)
Ta lại có: \(m_{NaOH}=\dfrac{5\%.240}{100\%}=12\left(g\right)\)
=> \(n_{NaOH}=\dfrac{12}{40}=0,3\left(mol\right)\)
a. PTHH: \(CuSO_4+2NaOH--->Cu\left(OH\right)_2\downarrow+Na_2SO_4\)
Ta thấy: \(\dfrac{0,1}{1}< \dfrac{0,3}{2}\)
Vậy NaOH dư, CuSO4 hết.
Theo PT: \(n_{Cu\left(OH\right)_2}=n_{Na_2SO_4}=n_{CuSO_4}=0,1\left(mol\right)\)
=> \(m_{Cu\left(OH\right)_2}=0,1.98=9,8\left(g\right)\)
b. Ta có: \(m_{dd_{Na_2SO_4}}=240+16-9,8=246,2\left(g\right)\)
Ta có: \(m_{Na_2SO_4}=0,1.142=14,2\left(g\right)\)
=> \(C_{\%_{Na_2SO_4}}=\dfrac{14,2}{246,2}.100\%=5,77\%\)
\(a,\left\{{}\begin{matrix}m_{H_2SO_4}=\dfrac{300\cdot9,8\%}{100\%}=29,4\left(g\right)\\m_{BaCl_2}=\dfrac{200\cdot26\%}{100\%}=52\left(g\right)\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}n_{H_2SO_4}=\dfrac{29,4}{98}=0,3\left(mol\right)\\n_{BaCl_2}=\dfrac{52}{208}=0,25\left(mol\right)\end{matrix}\right.\\ PTHH:H_2SO_4+BaCl_2\rightarrow2HCl+BaSO_4\downarrow\)
Vì \(\dfrac{n_{H_2SO_4}}{1}>\dfrac{n_{BaCl_2}}{1}\) nên sau phản ứng \(H_2SO_4\) dư
\(\Rightarrow n_{BaSO_4}=0,25\left(mol\right)\\ \Rightarrow a=m_{BaSO_4}=0,25\cdot233=58,25\left(g\right)\)
\(b,n_{HCl}=2n_{BaCl_2}=0,5\left(mol\right)\\ \Rightarrow m_{CT_{HCl}}=0,5\cdot36,5=18,25\left(g\right)\\ m_{dd_{HCl}}=300+200-58,25=441,75\left(g\right)\\ \Rightarrow C\%_{HCl}=\dfrac{18,25}{441,75}\cdot100\%\approx4,13\%\)
+mdd(H2SO4)=100*1.137=113.7(g)
=>mct(H2SO4)=113.7*20/100=22.74(g)
+nH2SO4=22.74/98=0.232(mol)
_mct(BaCl2)=400*5.2/100=20.8(g)
=>nBaCl2=20.8/208=0.1(mol)
-----------BaCl2+H2SO4=>BaSO4+2HCl
Trước pư.0.1......0.232......(mol)
Pứng.........0.1.......0.1..........0.1...
Sau pư.......0..........0.132..(mol)
=>H2SO4 còn dư 0.132 mol
_Sau phản ứng thu được kết tủa là BaSO4.
=>mBaSO4=0.1*233=23.3(g)
_Dung dịch B thu được gồm H2SO4 dư và HCl:
+nHCl=0.1*2=0.2(mol)
=>mHCl=0.2*36.5=7.3(g)
+nH2SO4 dư=0.132(mol)
=>mH2SO4 dư=0.132*98=12.936(g)
+mdd(sau)=113.7+400-23.3=490.4(g)
=>C%(HCl)=7.3*100/490.4=1.48%
=>C%(H2SO4 dư)=12.936*100/490.4=2.64%
Tính khối lượng kết tủa A
+mdd(H2SO4)=100*1.137=113.7(g)
=>mct(H2SO4)=113.7*20/100=22.74(g)
+nH2SO4=22.74/98=0.232(mol)
_mct(BaCl2)=400*5.2/100=20.8(g)
=>nBaCl2=20.8/208=0.1(mol)
-----------BaCl2+H2SO4=>BaSO4+2HCl
Trước pư.0.1......0.232......(mol)
Pứng.........0.1.......0.1..........0.1...
Sau pư.......0..........0.132..(mol)
=>H2SO4 còn dư 0.132 mol
_Sau phản ứng thu được kết tủa là BaSO4.
=>mBaSO4=0.1*233=23.3(g)
Tính nồng độ phần trăm của các chất trong dung dịch B
_Dung dịch B thu được gồm H2SO4 dư và HCl:
+nHCl=0.1*2=0.2(mol)
=>mHCl=0.2*36.5=7.3(g)
+nH2SO4 dư=0.132(mol)
=>mH2SO4 dư=0.132*98=12.936(g)
+mdd(sau)=113.7+400-23.3=490.4(g)
=>C%(HCl)=7.3*100/490.4=1.48%
=>C%(H2SO4 dư)=12.936*100/490.4=2.64%
\(n_{H_2SO_4}=\dfrac{200.19,6}{100.98}=0,4mol\\ CuO+H_2SO_4\rightarrow CuSO_4+H_2O\\ n_{CuSO_4\left(A\right)}=n_{CuO}=n_{H_2SO_4}=0,4mol\\ n_{Cu\left(OH\right)_2}=\dfrac{29,4}{98}=0,3mol\\ CuSO_4+2NaOH\rightarrow Cu\left(OH\right)_2+Na_2SO_4\\\Rightarrow\dfrac{0,4}{1}>\dfrac{0,3}{1}\Rightarrow CuSO_4.pư.không.hết\)
\(CuSO_4+2NaOH\rightarrow Cu\left(OH\right)_2+Na_2SO_4\)
0,3mol 0,6mol 0,3mol
\(m_{ddB}=0,4.80+200+0,6.40-29,4=226,6g\\ C_{\%Na_2SO_4\left(B\right)}=\dfrac{0,3.142}{226,6}\cdot100=18,8\%\)
Ta có: \(n_{H_2SO_4}=\dfrac{200.19,6\%}{98}=0,4\left(mol\right)\)
\(n_{Cu\left(OH\right)_2}=\dfrac{29,4}{98}=0,3\left(mol\right)\)
PT: \(CuO+H_2SO_4\rightarrow CuSO_4+H_2O\)
Dung dịch A gồm: CuSO4 và H2SO4 dư
\(H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O\)
\(CuSO_4+2NaOH\rightarrow Cu\left(OH\right)_2+Na_2SO_4\)
Đề có cho dữ kiện gì liên quan đến dd NaOH không bạn nhỉ?
Ta có: \(C_{\%_{KOH}}=\dfrac{m_{KOH}}{112}.100\%=56\%\)
=> mKOH = 62,72(g)
=> \(n_{KOH}=\dfrac{62,72}{56}=1,12\left(mol\right)\)
a. PTHH: 2KOH + MgCl2 ---> Mg(OH)2↓ + 2KCl
Theo PT: \(n_{Mg\left(OH\right)_2}=\dfrac{1}{2}.n_{KOH}=\dfrac{1}{2}.1,12=0,56\left(mol\right)\)
=> \(m_{Mg\left(OH\right)_2}=0,56.58=32,48\left(g\right)\)
b. Theo PT: \(n_{MgCl_2}=n_{Mg\left(OH\right)_2}=0,56\left(mol\right)\)
=> \(m_{MgCl_2}=0,56.95=53,2\left(g\right)\)
=> \(C_{\%_{MgCl_2}}=\dfrac{53,2}{200}.100\%=26,6\%\)
Ta có: \(\left\{{}\begin{matrix}n_{Ba\left(NO_3\right)_2}=\dfrac{26,1\%.10}{261}=0,01\left(mol\right)\\n_{H_2SO_4}=\dfrac{14,7\%.10}{98}=0,015\left(mol\right)\end{matrix}\right.\)
PTHH: \(Ba\left(NO_3\right)_2+H_2SO_4\rightarrow BaSO_4\downarrow+2HNO_3\)
ban đầu 0,01 0,015
phản ứng 0,01---------->0,01
sau phản ứng 0 0,005 0,01 0,02
`=>` \(m_{ddspư}=10+10-0,01.233=17,67\left(g\right)\)
`=>` \(C\%_{HNO_3}=\dfrac{0,02.63}{17,67}.100\%=7,13\%\)
\(Ba\left(NO_3\right)_2+H_2SO_4->BaSO_4+2HNO_3\)
\(n_{Ba\left(NO_3\right)_2}=\dfrac{10.26,1\%}{261}=0,01mol;n_{H_2SO_4}=\dfrac{10.14,7\%}{98}=0,015mol\)
Lập tỷ lệ \(\dfrac{n_{Ba\left(NO_3\right)_2}}{1}< \dfrac{n_{H_2SO_4}}{1}\) => Ba(NO3)2 hết
\(m_{BaSO_4}=233.0,01=2,33g\)
\(m_{HNO_3}=0,01.2.63=1,26g\)
=> \(C\%_{HNO_3}=\dfrac{1,26}{10+10-2,33}.100\%=7,13\%\)