Bài 7. Tính chất hóa học của bazơ - Hỏi đáp

c) \(PT:BaCl_2+H_2SO_4\rightarrow BaSO_4\downarrow+2HCl\)

0,2 0,2 (mol)

\(m_{BaSO_4}=0,2.233=46,6\left(g\right)\)

b) \(n_{H_2SO_4\left(\frac{1}{2}ddA\right)}=\frac{0,4}{2}=0,2\left(mol\right)\)

\(PT:Mg+H_2SO_4\rightarrow MgSO_4+H_2\)

0,2 0,2 (mol)

\(m_{Mg}=0,2.24=4,8\left(g\right)\)

2H2O -đp-> 2H2 + O2

NaCl + H2O -đpcmn-> NaOH + Cl2 + H2

H2 + Cl2 -to-> 2HCl

Fe + 2HCl --> FeCl2 + H2

FeCl2 + 2NaOH --> Fe(OH)2 + NaCl

\(\left(1\right)Fe+2H_2O\rightarrow Fe\left(OH\right)_2+H_2\uparrow\)

Điện phân nước ta được:

\(PTHH:2H_2O\underrightarrow{dp}2H_2+O_2\)

Cho Fe tác dụng với Oxi:

\(PTHH:2Fe+O_2\underrightarrow{t^o}2FeO\)

Cho Oxit td với nước:

\(\left(2\right)FeO+H_2O\rightarrow Fe\left(OH\right)_2\)

Cho Fe td với NaCl ở nhiệt độ cao

\(PTHH:Fe+2NaCl\underrightarrow{t^o}FeCl_2+2Na\)

Cho Na td với nước

\(PTHH:2Na+2H_2O\rightarrow2NaOH+H_2\)

Cho NaOH td với FeCl₂

\(\left(3\right)2NaOH+FeCl_2\rightarrow NaCl+Fe\left(OH\right)_2\)

PT: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

\(Fe+2HCl\rightarrow FeCl_2+H_2\)

a, Giả sử: \(\left\{{}\begin{matrix}n_{Al}=x\left(mol\right)\\n_{Fe}=y\left(mol\right)\end{matrix}\right.\)

⇒ 27x + 56y = 16,6 (1)

Ta có: \(n_{H_2}=\frac{11,2}{22,4}=0,5\left(mol\right)\)

Theo PT: \(\Sigma n_{H_2}=\frac{3}{2}n_{Al}+n_{Fe}=\frac{3}{2}x+y\left(mol\right)\)

\(\Rightarrow\frac{3}{2}x+y=0,5\left(2\right)\)

Từ (1) và (2) ⇒ x = y = 0,2 (mol)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Al}=\frac{0,2.27}{16,6}.100\%\approx32,53\%\\\%m_{Fe}\approx67,47\%\end{matrix}\right.\)

b, \(H_2+CuO\underrightarrow{t^o}Cu+H_2O\)

Ta có: \(n_{CuO}=\frac{16}{80}=0,2\left(mol\right)\)

Xét tỉ lệ: \(\frac{0,5}{1}>\frac{0,2}{1}\), ta được H₂ dư.

Theo PT: \(n_{Cu}=n_{CuO}=0,2\left(mol\right)\)

\(\Rightarrow m_{Cu}=0,2.64=12,8\left(g\right)\)

Bạn tham khảo nhé!

a, Giả sử CT muối sắt cần tìm là Fe_x(SO₄)_y

Theo đầu bài, ta có: \(\frac{56x}{56x+96y}=0,28\)

\(\Rightarrow3x=2y\)

\(\Rightarrow\frac{x}{y}=\frac{2}{3}\)

Vậy: Muối sắt đó là Fe₂(SO₄)₃

b, PT: \(Fe_2\left(SO_4\right)_3+6KOH\rightarrow3K_2SO_4+2Fe\left(OH\right)_{3\downarrow}\)

Kết tủa Y là Fe(OH)₃.

Theo PT: \(\left\{{}\begin{matrix}n_{Fe_2\left(SO_4\right)_3}=\frac{1}{6}n_{KOH}=0,07\left(mol\right)\\n_{Fe\left(OH\right)_3\downarrow}=\frac{1}{3}n_{KOH}=0,14\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=0,07\left(mol\right)\\m_Y=m_{Fe\left(OH\right)_3}=0,14.107=14,98\left(g\right)\end{matrix}\right.\)

Bạn tham khảo nhé!

a) CTHH: \(Fe_x\left(SO_4\right)_y\)

=> \(\frac{56x}{96y}=\frac{28\%}{72\%}\) => \(\frac{x}{y}=\frac{2}{3}\) => \(\left\{{}\begin{matrix}x=2\\y=3\end{matrix}\right.\)

=> CTHH: \(Fe_2\left(SO_4\right)_3\)

b)

PTHH: \(Fe_2\left(SO_4\right)_3+6KOH\rightarrow2Fe\left(OH\right)_3\downarrow+3K_2SO_4\)

________ 0,07 <--------- 0,42 --------> 0,14_____________(mol)

=> \(\left\{{}\begin{matrix}x=0,07\left(mol\right)\\m_Y=m_{Fe\left(OH\right)_3}=0,14.107=14,98\left(g\right)\end{matrix}\right.\)

\(Đặt\) : \(n_{hh}=1\left(mol\right)\\ n_{CO_2}=x\left(mol\right)\Rightarrow n_{SO_2}=1-x\left(mol\right)\)

\(\overline{M}=26\cdot1\cdot2=52\left(g\right)\)

\(\Leftrightarrow44x+64\left(1-x\right)=52\Leftrightarrow x=0.6\)

\(\%CO_2=60\%\\ \%SO_2=40\%\)

\(\frac{n_{CO_2}}{n_{SO_2}}=\frac{0.6}{0.4}=\frac{3}{2}\)\(\Leftrightarrow3n_{SO_2}-2n_{CO_2}=0\left(1\right)\)

\(n_{hh}=n_{SO_2}+n_{CO_2}=\frac{11.2}{22.4}=0.5\left(2\right)\)

\(\left(1\right),\left(2\right)\Leftrightarrow\left\{{}\begin{matrix}n_{SO_2}=0.2\\n_{CO_2}=0.3\end{matrix}\right.\)

\(Ba\left(OH\right)_2+CO_2\Rightarrow BaCO_3+H_2O\\Ba\left(OH\right)_2+SO_2\Rightarrow BaSO_3+H_2O\)

\(m_{KT}=0.3\cdot197+0.2\cdot217=102.5\left(g\right)\)

a) Gọi số mol CO2 và SO2 là a,b (mol)

Ta có: \(\overline{M_X}=\frac{44a+64b}{a+b}=26.2\) => a = 1,5b

\(=>\left\{{}\begin{matrix}\%n_{CO_2}=\frac{a}{a+b}.100\%=\frac{1,5b}{1,5b+b}.100\%=60\%\\\%n_{SO_2}=\frac{b}{a+b}.100\%=\frac{b}{1,5b+b}.100\%=40\%\end{matrix}\right.\)

b) \(n_X=\frac{11,2}{22,4}=0,5\left(mol\right)\)

Mà a = 1,5b

=> \(\left\{{}\begin{matrix}a=0,3\left(mol\right)\\b=0,2\left(mol\right)\end{matrix}\right.\)

PTHH: \(Ba\left(OH\right)_2+CO_2\rightarrow BaCO_3+H_2O\)

__________________ 0,3 --------> 0,3_______(mol)

\(Ba\left(OH\right)_2+SO_2\rightarrow BaSO_3+H_2O\)

____________0,2 ------> 0,2_________(mol)

=> \(\left\{{}\begin{matrix}m_{BaCO_3}=0,3.197=59,1\left(g\right)\\m_{BaSO_3}=0,2.217=43,4\left(g\right)\end{matrix}\right.\)

=> \(m_{KT}=59,1+43,4=102,5\left(g\right)\)

\(V_{ddNaOH}=\frac{240}{1.2}=200\left(ml\right)=0.2\left(l\right)\\ n_{NaOH}=0.2\cdot1=0.2\left(mol\right)\)

\(C_{M_{HCl}}=\frac{C\%_{HCl}\cdot10D}{M}=\frac{14.6\cdot10\cdot1.1}{36.5}=4.4M\\ n_{HCl}=4.4\cdot0.05=0.22\left(mol\right)\)

\(.....NaOH+HCl\rightarrow NaCl+H_2O\)

\(Bđ:\) \(0.2............0.22\)

\(Pư:\) \(0.2.............0.2.........0.2\)

\(Kt:\) \(0...............0.02........0.2\)

\(m_{dd}=240+50\cdot1.1=295\left(g\right)\)

\(m_{NaCl}=0.2\cdot58.5=11.7\left(g\right)\\ C\%_{NaCl}=\frac{11.7}{295}\cdot100\%=3.96\%\)

\(m_{HCl}dư=0.02\cdot36.5=0.73\left(g\right)\\ C\%_{HCl}dư=\frac{0.73}{295}\cdot100\%=0.24\%\)

\(Al_2O_3+6HCl\rightarrow2AlCl_3+3H_2\)

\(\frac{1}{300}.......0.02\)

\(m_{Al_2O_3}=\frac{102}{300}=0.34\left(g\right)\)

\(n_{H_2}=\frac{3.36}{22.4}=0.15\left(mol\right)\)

\(n_{AlCl_3}=\frac{40.05}{133.5}=0.3\left(mol\right)\)

\(Al+NaOH+H_2O\rightarrow NaAlO_2+\frac{3}{2}H_2\)

\(0.1...................................................0.15\)

\(2NaOH+Al_2O_3\rightarrow2NaAlO_2+H_2O\)

\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

\(0.1.......................0.1\)

\(Al_2O_3+6HCl\rightarrow2AlCl_3+3H_2\)

\(0.1.......................0.3-0.1\)

\(m=m_{Al}+m_{Al_2O_3}=0.1\cdot27+0.1\cdot102=12.9\left(g\right)\)

Gọi số mol Fe(OH)3, Mg(OH)2 là a,b (mol)

PTHH: \(2Fe\left(OH\right)_3\underrightarrow{t^o}Fe_2O_3+3H2O\)

________a ------------> 0,5a____________(mol)

\(Mg\left(OH\right)_2\underrightarrow{t^o}MgO+H_2O\)

b-----------------> b_________(mol)

=> \(\left\{{}\begin{matrix}107a+58b=27,2\\160.0,5a+40b=27,2-7,2\end{matrix}\right.\) => \(\left\{{}\begin{matrix}a=0,2\left(mol\right)\\b=0,11\left(mol\right)\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}m_{Fe\left(OH\right)_3}=0,2.107=21,4\\m_{Mg\left(OH\right)_2}=0,1.58=5,8\left(g\right)\\m_{Fe_2O_3}=160.0,5.0,2=16g\\m_{MgO}=40.0,1=4\left(g\right)\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}\%m_{Fe\left(OH\right)_3}=\frac{21,4}{27,2}.100\%=78,68\%\\\%m_{Mg\left(OH\right)_2}=\frac{5,8}{27,2}.100\%=21,32\%\end{matrix}\right.\)

\(a.\)

\(Đăt:n_{Mg}=x\left(mol\right),n_{Fe}=y\left(mol\right)\)

\(Mg+H_2SO_4\rightarrow MgSO_4+H_2\\ Fe+H_2SO_4\rightarrow FeSO_4+H_2\\ MgSO_4+2NaOH\rightarrow Mg\left(OH\right)_2+Na_2SO_4\\ FeSO_4+2NaOH\rightarrow Fe\left(OH\right)_2+Na_2SO_4\\ Mg\left(OH\right)_2\underrightarrow{t^0}MgO+H_2O\\ 4Fe\left(OH\right)_2+O_2\underrightarrow{t^0}2Fe_2O_3+4H_2O\)

\(\left\{{}\begin{matrix}m_{hh}=24x+56y=17.6\left(g\right)\\m_{kt}=58x+90y=31.2\left(g\right)\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0.15\\y=0.25\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}m_{Mg}=3.6g\\m_{Fe}=14g\end{matrix}\right.\)

\(m_{H_2SO_4}=\left(0.15+0.25\right)\cdot98=39.2\left(g\right)\)

\(BTMg:n_{Mg}=n_{MgO}=0.15\left(mol\right)\Rightarrow m_{MgO}=6\left(g\right)\)

\(BTFe:n_{Fe_2O_3}=\frac{n_{Fe}}{2}=\frac{0.25}{2}=0.125\left(mol\right)\)

\(\Leftrightarrow m_{Fe_2O_3}=0.125\cdot160=20\left(g\right)\)

\(\overline{M}=\frac{m_{CO_2}+m_A}{n_{CO_2}+n_A}=\frac{0.2\cdot44+0.3A}{0.2+0.3}=\frac{8.8+0.3A}{0.5}\)

\(d_{\frac{\overline{M}}{H_2}}=17.2\Rightarrow\overline{M}=17.2\cdot2=34.4\)

\(\Leftrightarrow\frac{8.8+0.3A}{0.5}=34.4\\ \Leftrightarrow A=28\\ Vậy:A:N_2\)

b) \(n_X=\frac{11.2}{22.4}=0.5\left(mol\right)\)

\(\frac{n_{CO_2}}{n_{N_2}}=\frac{0.2}{0.3}=\frac{2}{3}\Leftrightarrow2n_{N_2}-3n_{CO_2}=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}n_{N_2}=0.3\left(mol\right)\\n_{CO_2}=0.2\left(mol\right)\end{matrix}\right.\)

\(CO_2+Ca\left(OH\right)_2\Rightarrow CaCO_3+H_2O\\ n_{CO_2}=n_{CaCO_3}=0.2\left(mol\right)\)

\(\Rightarrow m_{CaCO_3}=0.2\cdot100=20\left(g\right)\)

a, Theo đầu bài, ta có:

\(d_{X/H_2}=17,2\Rightarrow\overline{M_X}=17,2.2=34,4\)

Hay: \(\frac{0,2.44+0,3.M_A}{0,2+0,3}=34,4\)

\(\Rightarrow M_A=28\)

Vậy: Khí A là N₂.

b, Ta có: \(n_{hhkX}=\frac{11,2}{22,4}=0,5\left(mol\right)=n_{CO_2}+n_{N_2}\left(0,2+0,3\right)\)

PT: \(CO_2+Ca\left(OH\right)_{2\left(dư\right)}\rightarrow CaCO_3+H_2O\)

Theo PT: \(n_{CaCO_3\downarrow}=n_{CO_2}=0,2\left(mol\right)\)

\(\Rightarrow m_{CaCO_3\downarrow}=0,2.100=20\left(g\right)\)

Bạn tham khảo nhé!

\(A:\left\{{}\begin{matrix}Mg^{2+}:0.2x\left(mol\right)\\Al^{3+}:0.4y\left(mol\right)\\SO_4^{2-}:0.2x+0.6y\left(mol\right)\end{matrix}\right.\)

Vì thêm NaOH đến dư => Al(OH)₃ bị hòa tan hoàn toàn

\(BTMg:n_{MgO}=n_{Mg^{2+}}=\frac{8}{40}=0.2\left(mol\right)\)

\(\Rightarrow0.2x=0.2\Rightarrow x=1\)

\(n_{BaSO_4}=\frac{116.5}{233}=0.5\left(mol\right)\)

\(Ba^{2+}+SO_4^{2-}\rightarrow BaSO_4\)

\(\Rightarrow n_{SO_4^{2-}}=n_{BaSO_4}=0.5\left(mol\right)\)

\(\Leftrightarrow0.2x+0.6y=0.5\\ \Rightarrow y=\frac{0.5-0.2}{0.6}=0.5\)

\(n_{HCl}=0.2\left(mol\right)\)

\(NaOH+HCl\rightarrow NaCl+H_2O\)

\(0.2............0.2..........0.2\)

\(V_{ddNaOH}=\frac{0.2}{1.5}=\frac{2}{15}=0.133\left(l\right)\)

\(m_{NaCl}=0.2\cdot58.5=11.7\left(g\right)\)

\(C_{M_{NaCl}}=\frac{0.2}{0.2+\frac{2}{15}}=0.6M\)

a, PTHH : \(NaOH+HCl\rightarrow NaCl+H_2O\)

\(n_{HCl}=C_M.V=0,2.1=0,2\left(mol\right)\)

- Theo PTHH : \(n_{NaOH}=n_{HCl}=0,2\left(mol\right)\)

=> \(V_{NaOH}=\frac{n}{C_M}=\frac{0,2}{1,5}=\frac{2}{15}\left(l\right)\)

b, - Theo PTHH : \(n_{NaCl}=n_{HCl}=0,2\left(mol\right)\)

=> \(m_{NaCl}=n.M=0,2.58,5=11,7\left(g\right)\)

Ta có : \(V_{dd}=0,2+\frac{2}{15}=\frac{1}{3}\left(l\right)\)

=> \(C_M=\frac{n}{V}=\frac{0,2}{\frac{1}{3}}=0,6\left(M\right)\)

PT: \(2HCl+Ca\left(OH\right)_2\rightarrow CaCl_2+2H_2O\)

a, Ta có: \(n_{HCl}=\frac{18,25}{36,5}=0,5\left(mol\right)\)

\(n_{Ca\left(OH\right)_2}=0,2.0,5=0,1\left(mol\right)\)

Xét tỉ lệ: \(\frac{0,5}{2}>\frac{0,1}{1}\), ta được HCl dư.

Các chất sau phản ứng gồm: HCl dư và CaCl₂.

Theo PT: \(\left\{{}\begin{matrix}n_{CaCl_2}=n_{Ca\left(OH\right)_2}=0,1\left(mol\right)\\n_{HCl\left(pư\right)}=2n_{Ca\left(OH\right)_2}=0,2\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow n_{HCl\left(dư\right)}=0,3\left(mol\right)\)

\(\Rightarrow m_{CaCl_2}=0,1.111=11,1\left(g\right)\)

\(m_{HCl\left(dư\right)}=0,3.36,5=10,95\left(g\right)\)

b, Vì: D = 1,25 g/ml

⇒ m _{dd Ca(OH)2} = 1,25.200 = 250 (g)

Ta có: m _{dd sau pư} = m_HCl + m _{dd Ca(OH)2} = 18,25 + 250 = 268,25 (g)

\(\Rightarrow\left\{{}\begin{matrix}C\%_{CaCl_2}=\frac{11,1}{268,25}.100\%\approx4,138\%\\C\%_{HCl\left(dư\right)}=\frac{10,95}{268,25}.100\%\approx4,082\%\end{matrix}\right.\)

c, Ta có: V _{dd sau pư} = V _{dd Ca(OH)2} = 200 (ml) = 0,2 (lít)

\(\Rightarrow\left\{{}\begin{matrix}C_{M_{CaCl_2}}=\frac{0,1}{0,2}=0,5M\\C_{M_{HCl\left(dư\right)}}=\frac{0,3}{0,2}=1,5M\end{matrix}\right.\)

Bạn tham khảo nhé!

n_Ba(OH)2= 0,15 (mol)

n_NaOH= 0,3 (mol)

n_BaCO3 = 0,1 (mol)

Các pt xảy ra theo thứ tự:

Ba(OH)₂ + CO₂ \(\rightarrow\) BaCO₃ + H₂O (1)

2NaOH + CO₂ \(\rightarrow\) Na₂CO₃ + H₂O (2)

nếu CO₂ dư:

Na₂CO₃ + CO₂ + H₂O \(\rightarrow\) 2NaHCO₃ (3)

BaCO₃ + CO₂ + H₂O \(\rightarrow\) Ba(HCO₃)₂ (4)

+ TH₁: CO₂ pư vừa đủ với dd hh bazơ

Theo pt(1) n_BaCO3 = n_Ba(OH)2 = 0,15 \(\ne\) 0,1 (mol) (loại)

+ TH₂: CO₂ pư thiếu so với dd hh bazơ \(\Rightarrow\) xảy ra pt (1)

Theo pt(1) n_CO2 = n_BaCO3= 0,1 (mol)

V_CO2= 0,1 . 22,4 = 2,24 (l)

+ TH₃: CO₂ pư dư so với dd hh bazơ \(\Rightarrow\) xảy ra pt(1), (2), (3), (4)

Ba(OH)₂ + CO₂ \(\rightarrow\) BaCO₃ \(\downarrow\) + H₂O

0,15 \(\rightarrow\) 0,15 \(\rightarrow\) 0,15 (mol)

2NaOH + CO₂ \(\rightarrow\) Na₂CO₃ + H₂O

0,3 \(\rightarrow\) 0,15 \(\rightarrow\) 0,15 (mol)

Na₂CO₃ + CO₂ + H₂O \(\rightarrow\) 2NaHCO₃

0,15 \(\rightarrow\) 0,15 (mol)

BaCO₃ + CO₂ + H₂O \(\rightarrow\) Ba(HCO₃)₂

(0,15 - 0,1) \(\rightarrow\) 0,05 (mol)

V_CO2 = (0,15 + 0,15 + 0,15 + 0,05).22,4= 11,2 (l)

nBa(OH)2= 0,15 (mol)

nNaOH= 0,3 (mol)

nBaCO3 = 0,1 (mol)

Các pt xảy ra theo thứ tự:

Ba(OH)2 + CO2 BaCO3 + H2O (1)

2NaOH + CO2 Na2CO3 + H2O (2)

nếu CO2 dư:

Na2CO3 + CO2 + H2O 2NaHCO3 (3)

BaCO3 + CO2 + H2O Ba(HCO3)2 (4)

+ TH1: CO2 pư vừa đủ với dd hh bazơ

Theo pt(1) nBaCO3 = nBa(OH)2 = 0,15 0,1 (mol) (loại)

+ TH2: CO2 pư thiếu so với dd hh bazơ xảy ra pt (1)

Theo pt(1) nCO2 = nBaCO3= 0,1 (mol)

VCO2= 0,1 . 22,4 = 2,24 (l)

+ TH3: CO2 pư dư so với dd hh bazơ xảy ra pt(1), (2), (3), (4)

Ba(OH)2 + CO2 BaCO3 + H2O

0,15 0,15 0,15 (mol)

2NaOH + CO2 Na2CO3 + H2O

0,3 0,15 0,15 (mol)

Na2CO3 + CO2 + H2O 2NaHCO3

0,15 0,15 (mol)

BaCO3 + CO2 + H2O Ba(HCO3)2

(0,15 - 0,1) 0,05 (mol)

VCO2 = (0,15 + 0,15 + 0,15 + 0,05).22,4= 11,2 (l)