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\(oxit.bazo\\ K_2O:kalioxit\\ Na_2O:natrioxit\\ Fe_2O_3:sắt\left(III\right)oxit\\ FeO:sắt\left(II\right)oxit\\ SiO_2:silcoxit\)
\(oxit.axit:\\ CO_2:cacbonic\\ SO_3:lưu.huỳnh.tri.oxit\\ SO_2:lưu.huỳnh.đi.oxit\)
\(bazo\\ Mg\left(OH\right)_2:magiehidroxit\\ NaOH:natrihidroxit\\ Ca\left(OH\right)_2:canxihidroxit\)
\(axit\\ HNO_3:axitnitric\\ H_2CO_3:axitcacbonic\\ H_2SO_4:axitsunfuric\\ H_2S:axitsunfua\)
\(muối\\ CuCl_2:đồng\left(II\right)clorua\\ CaCO_3:canxicacbonat\\ Fe_2\left(SO_4\right)_3:sắt\left(III\right)sunfat\\ K_3PO_4:kaliphotphat\\ BaSO_3:barisunfit\\ ZnSO_4:kẽmsunfat\\ Al\left(NO_3\right)_3:nhômnitrat\\ NaHCO_3:natrihidrocacbonat\\ Ca\left(HCO_3\right)_2:canxihidrocacbonat\)
Ta có: \(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)
\(Zn+2HCl\rightarrow ZnCl_2+H_2\)
Theo PTHH:
\(n_{H_2}=n_{Zn}=0,1\left(mol\right)\)
\(\Rightarrow V_{H_2}=0,1\cdot22,4=2,24\left(l\right)\)
@Phong em nhớ sách mới lớp 8 thì 1 mol ở 25 độ C và áp suất 1 bar là 24,79 lít. Còn 22,4 lít là sách cũ tính theo 0 độ C và 1 Asmosphere.
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\(a,C\%_A=\dfrac{12,5}{12,5+87,5}.100\%=12,5\%\)
\(b,PTHH:\)
\(CuSO_4+2NaOH\rightarrow Cu\left(OH\right)_2\downarrow+Na_2SO_4\)
trc p/u : 0,05 0,15
p/u : 0,05 0,1 0,05 0,05
sau: 0 0,05 0,05 0,05 (mol)
-> sau p/ư NaOH dư .
\(n_{NaOH}=\dfrac{40.15\%}{40}=0,15\left(mol\right)\)
\(n_{CuSO_4}=\dfrac{12,5}{250}=0,05\left(mol\right)\)
\(m_{Cu\left(OH\right)_2}=0,05.98=4,9\left(g\right)\)
\(m_{ddB}=100+40=140\left(g\right)\)
\(C\%_B=\dfrac{4,9}{140}.100\%=3,5\%\)
PT: \(MgO+H_2SO_4\rightarrow MgSO_4+H_2O\)
\(FeO+H_2SO_4\rightarrow FeSO_4+H_2O\)
Gọi: \(\left\{{}\begin{matrix}n_{MgO}=x\left(mol\right)\\n_{FeO}=y\left(mol\right)\end{matrix}\right.\) ⇒ 40x + 72y = 4,88 (1)
Ta có: \(n_{H_2SO_4}=0,2.0,45=0,09\left(mol\right)\)
Theo PT: \(n_{H_2SO_4}=n_{MgO}+n_{FeO}=x+y=0,09\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}x=0,05\left(mol\right)\\y=0,04\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{MgO}=\dfrac{0,05.40}{4,88}.100\%\approx40,98\%\\\%m_{FeO}\approx59,02\%\end{matrix}\right.\)
a, PT: \(CuO+2HCl\rightarrow CuCl_2+H_2O\)
\(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\)
Gọi: \(\left\{{}\begin{matrix}n_{CuO}=x\left(mol\right)\\n_{Fe_2O_3}=y\left(mol\right)\end{matrix}\right.\) ⇒ 80x + 160y = 11,2 (1)
Ta có: \(m_{HCl}=146.10\%=14,6\left(g\right)\Rightarrow n_{HCl}=\dfrac{14,6}{36,5}=0,4\left(mol\right)\)
Theo PT: \(n_{HCl}=2n_{CuO}+6n_{Fe_2O_3}=2x+6y=0,4\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}x=0,02\left(mol\right)\\y=0,06\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{CuO}=\dfrac{0,02.80}{11,2}.100\%\approx14,29\%\\\%m_{Fe_2O_3}\approx85,71\%\end{matrix}\right.\)
b, PT: \(CuO+H_2SO_4\rightarrow CuSO_4+H_2O\)
\(Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O\)
Theo PT: \(n_{H_2SO_4}=n_{CuO}+3n_{Fe_2O_3}=0,2\left(mol\right)\)
\(\Rightarrow m_{H_2SO_4}=0,2.98=19,6\left(g\right)\Rightarrow m_{ddH_2SO_4}=\dfrac{19,6}{4,9\%}=400\left(g\right)\)
Quá trình:
\(Fe_x^{^{+\dfrac{2y}{x}}}\rightarrow xFe^{+3}+\left(3x-2y\right)|\times4\)
\(N^{+5}+3e\rightarrow N^{+2}|\times\left(3x-2y\right)\)
\(N^{+5}+1e\rightarrow N^{+4}|\times\left(3x-2y\right)\)
PT: \(4Fe_xO_y+\left(18x-4y\right)HNO_3\rightarrow4xFe\left(NO_3\right)_3+\left(3x-2y\right)NO+\left(3x-2y\right)NO_2+\left(9x-2y\right)H_2O\)
\(a,PTHH:Zn+2HCl\rightarrow ZnCl_2+H_2\\ n_{HCl}=\dfrac{150.3,65\%}{36,5}=0,15\left(mol\right)\\ n_{Zn}=n_{H_2}=n_{ZnCl_2}=\dfrac{0,15}{2}=0,075\left(mol\right)\\ b,m_{Zn}=0,075.65=4,875\left(g\right)\\c,m_{ddsau}=4,875+150-0,075.2=154,725\left(g\right)\\ m_{ZnCl_2}=0,075.136=10,2\left(g\right)\\c, C\%_{ddZnCl_2}=\dfrac{10,2}{154,725}.100\%\approx6,592\%\\ V_{ddsau}=V_{ddHCl}=\dfrac{150}{1,2}=125\left(ml\right)=0,125\left(l\right)\\ C_{MddZnCl_2}=\dfrac{0,075}{0,125}=0,6\left(M\right)\)