Dẫn 8,96 lít khí SO2 (đktc) vào 200 gam dd NaOH 18%. Tính nồng độ % các chất sau phản ứng
Dẫn 8,96 lít khí SO2 (đktc) vào 200 gam dd NaOH 18%. Tính nồng độ % các chất sau phản ứng
\(n_{SO_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\\ m_{NaOH}=\dfrac{200.18}{100}=36\left(g\right)\\ \rightarrow n_{NaOH}=\dfrac{36}{40}=0,9\left(mol\right)\)
Xét \(T=\dfrac{0,9}{0,4}=2,5\) => Tạo muối Na2SO3 và NaOH dư
PTHH: 2NaOH + SO2 ---> Na2SO3 + H2O
0,8<----0,4--------->0,4
\(m_{dd}=0,4.64+200=225,6\left(g\right)\\ \rightarrow\left\{{}\begin{matrix}C\%_{Na_2SO_3}=\dfrac{0,4.126}{225,6}.100\%=22,34\%\\C\%_{NaOH\left(dư\right)}=\dfrac{\left(0,9-0,8\right).40}{225,6}.100\%=1,77\%\end{matrix}\right.\)
Bài 2:
\(n_{CuSO_4}=n_{CuSO_4.2H_2O}=\dfrac{25}{196}\left(mol\right)\\C_{MddCuSO_4}=\dfrac{\dfrac{25}{196}}{0,2}\approx0,638\left(M\right) \)
Bài 3:
\(m_{CuSO_4}=\dfrac{160}{196}.25=\dfrac{1000}{49}\left(g\right)\\ C\%_{ddCuSO_4}=\dfrac{\dfrac{1000}{49}}{200}.100\approx10,204\%\)
Giải chi tiết.
Câu 1 :
\(2KMnO_4+16HCl_{\left(đ\right)}\underrightarrow{^{t^0}}2KCl+2MnCl_2+5Cl_2+8H_2O\)
\(H_2+Cl_2\underrightarrow{^{as}}2HCl\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
\(FeCl_2+2NaOH\rightarrow Fe\left(OH\right)_2+2NaCl\)
Câu 2 :
\(n_{Mg}=a\left(mol\right),n_{Zn}=b\left(mol\right)\)
\(m_{hh}=24a+65b=14.55\left(g\right)\left(1\right)\)
\(n_{H_2}=\dfrac{7.84}{22.4}=0.35\left(mol\right)\)
\(Mg+H_2SO_4\rightarrow MgSO_4+H_2\)
\(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)
\(n_{H_2}=a+b=0.35\left(mol\right)\left(2\right)\)
\(\left(1\right),\left(2\right):a=0.2,b=0.15\)
\(\%m_{Mg}=\dfrac{0.2\cdot24}{14.55}\cdot100\%=33.98\%\)
\(\%m_{Zn}=100-33.98=67.01\%\)
Câu 3 :
\(m_{FeS_2}=187.5\cdot80\%=150\left(kg\right)\)
\(n_{FeS_2}=\dfrac{150\cdot10^3}{120}=1250\left(mol\right)\)
\(n_{H_2SO_4}=2n_{FeS_2}=1250\cdot2=2500\left(mol\right)\)
\(n_{H_2SO_4\left(tt\right)}=2500\cdot90\%=2250\left(mol\right)\)
\(m_{dd_{H_2SO_4}}=\dfrac{2250\cdot98\cdot100}{98}=225000\left(g\right)\)
\(V_{dd_{H_2SO_4}}=\dfrac{225000}{1.84}=122282\left(ml\right)=122.282\left(l\right)\)
Giải chi tiết.
a) Đặt \(\left\{{}\begin{matrix}n_{Mg}=a\left(mol\right)\\n_{Fe}=b\left(mol\right)\end{matrix}\right.\) \(\Rightarrow24a+56b=22,8\) (1)
Ta có: \(n_{SO_2}=\dfrac{15,68}{22,4}=0,7\left(mol\right)\)
Bảo toàn electron: \(2a+3b=0,7\cdot2=1,4\) (2)
Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}a=0,25\\b=0,3\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\%m_{Mg}=\dfrac{0,25\cdot24}{22,8}\cdot100\%\approx26,32\%\\\%m_{Fe}=73,68\%\end{matrix}\right.\)
b) PTHH: \(2NaOH+H_2SO_{4\left(dư\right)}\rightarrow Na_2SO_4+2H_2O\)
Theo PTHH: \(n_{H_2SO_4\left(dư\right)}=\dfrac{1}{2}n_{NaOH}=\dfrac{0,4\cdot3}{2}=0,6\left(mol\right)\)
Ta có: \(\left\{{}\begin{matrix}n_{Fe_2\left(SO_4\right)_3}=\dfrac{1}{2}n_{Fe}=0,15\left(mol\right)\\n_{MgSO_4}=n_{Mg}=0,25\left(mol\right)\end{matrix}\right.\)
Bảo toàn nguyên tố: \(n_{H_2SO_4\left(p/ứ\right)}=n_{MgSO_4}+3n_{Fe_2\left(SO_4\right)_3}+n_{SO_2}=1,4\left(mol\right)\)
\(\Rightarrow\Sigma n_{H_2SO_4}=1,4+0,6=2\left(mol\right)\) \(\Rightarrow m_{ddH_2SO_4}=\dfrac{2\cdot98}{98\%}=200\left(g\right)\)
Từ câu 08 đến câu 10.
Bài 8 :
Coi Oxit sắt gồm Fe(x mol) và O(y mol)
=> 56x + 16y = 36(1)
n SO2 = 5,6/22,4 = 0,25(mol)
Bảo toàn electron :
3n Fe = 2n SO2 + 2n O
<=> 3a - 2b = 0,5(2)
Từ(1)(2) suy ra a = 0,5 ; b = 0,5
n Fe / n O = 0,5/0,5 = 1/1
Vậy oxit sắt là FeO
Bài 8 :
CT : FexOy
\(n_{SO_2}=\dfrac{5.6}{22.4}=0.25\left(mol\right)\)
\(xFe^{\dfrac{2y}{x}}\rightarrow xFe^{3+}+\left(3x-2y\right)e\)
\(S^{+6}+2e\rightarrow S^{+4}\)
\(BTe:\)
\(n_{Fe_xO_y}=\dfrac{0.5}{3x-2y}\left(mol\right)\)
\(M=\dfrac{36}{\dfrac{0.5}{3x-2y}}=72\left(3x-2y\right)\left(\dfrac{g}{mol}\right)\)
\(\Leftrightarrow56x+16y=72\cdot\left(3x-2y\right)\)
\(\Leftrightarrow160x=160y\)
\(\Leftrightarrow\dfrac{x}{y}=\dfrac{1}{1}\)
\(CT:FeO\)
Câu 9 :
n S = 3,2/32 = 0,1(mol)
n SO2 = 13,44/22,4 = 0,6(mol)
Bảo toàn e với cả quá trình :
\(Fe^0 \to Fe^{3+} + 3e\\ S^0 \to S^{+6} + 6e\\ S^{+6} + 2e \to S^{+4}\)
Ta có: 3n Fe + 6n S = 2n SO2
<=> n Fe = (0,6.2 - 0,1.6)/3 = 0,2(mol)
<=> m = 0,2.56 = 11,2(gam)
Từ câu 06 đến câu 07.
Bài 6 :
a) n Mg = a(mol) ; n Fe = b(mol) => 24a + 56b = 6,8(1)
Mg + 2HCl → MgCl2 + H2
Fe + 2HCl → FeCl2 + H2
Theo PTHH : n H2 = a + b = 3,36/22,4 = 0,15(2)
(1)(2) => a = 0,05 ; b = 0,1
%m Mg = 0,05.24/6,8 .100% = 17,65%
%m Fe = 100% -17,65% = 82,35%
b) Bảo toàn e :
2n Mg + 3n Fe = 2n SO2
=> n SO2 = (0,05.2 + 0,1.3)/2 = 0,2 mol
=> V SO2 = 0,2.22,4 = 4,48(lít)
Bài 7 :
\(n_{H_2SO_4}=\dfrac{100\cdot10^6\cdot98}{100\cdot98}=10^6\left(mol\right)\)
\(n_{FeS_2}=\dfrac{10^6}{2}=5\cdot10^5\left(mol\right)\)
\(n_{FeS_2\left(tt\right)}=\dfrac{5\cdot10^5\cdot120}{75\%}=80\left(tấn\right)\)
Câu 7 :
n H2SO4 = 100.1000.98%/98 = 1000(kmol)
Bảo toàn nguyên tố với S :
n FeS2 pư = 1/2 n H2SO4 = 500(kmol)
n FeS2 đã dùng = 500/75% = 2000/3(kmol)
m FeS2 = 120 . 2000/3 = 80 000(kg)
m pirit = 80 000/10% = 800 000(kg)
Từ câu 04 đến câu 05.
Bài 5 :
n Cu = a(mol) ; n Al = b(mol) ; n Mg = c(mol)
=> 64a + 27b + 24c = 11,5(1)
2Al + 6HCl $\to$ 2AlCl3 + 3H2
Mg + 2HCl $\to$ MgCl2 + H2
Theo PTHH :
n H2 = 1,5b + c = 5,6/22,4 = 0,25(2)
Bảo toàn electron :
2n SO2 = 2n Cu
2.0,1 = 2a (3)
Từ (1)(2)(3) suy ra a = 0,1 ; b = 0,1 ; c = 0,1
Vậy :
m Cu = 0,1.64 = 6,4 gam
m Al = 0,1.27 = 2,7 gam
m Mg = 0,1.24 = 2,4 gam
Từ câu 01 đến câu 02.
Bài 1 :
\(n_{Mg}=a\left(mol\right),n_{Zn}=b\left(mol\right)\)
\(m_{hh}=24a+65b=31.5\left(g\right)\left(1\right)\)
\(n_{H_2}=\dfrac{17.92}{22.4}=0.8\left(mol\right)\)
\(Mg+H_2SO_4\rightarrow MgSO_4+H_2\)
\(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)
\(n_{H_2}=a+b=0.8\left(mol\right)\left(2\right)\)
\(\left(1\right),\left(2\right):a=0.5,b=0.3\)
\(m_{Mg}=0.5\cdot24=12\left(g\right)\)
\(m_{Zn}=0.3\cdot19.5\left(g\right)\)
\(n_{H_2SO_4}=n_{H_2}=0.8\left(mol\right)\)
\(C\%H_2SO_4=\dfrac{0.8\cdot98}{300}\cdot100\%=26.13\%\)
Từ câu 01 đến 05( chuỗi hóa học).
1/
\(4FeS_2 + 11O_2 \xrightarrow{t^o} 2Fe_2O_3 + 8SO_2\\ SO_2 + 2H_2S \to 3S + 2H_2O\\ S + H_2 \xrightarrow{t^o,xt} H_2S\\ H_2S + 4Br_2 + 4H_2O \to 8HBr + H_2SO_4\\ H_2SO_4 + NaCl \xrightarrow{t^o} NaHSO_4 + HCl\\ MnO_2 + 4HCl \to MnCl_2 + Cl_2 + 2H_2O\\ 2Na +C l_2 \xrightarrow{t^o} 2NaCl\\ NaCl + AgNO_3 \to AgCl + NaNO_3\)
2/
\(S + H_2 \xrightarrow{t^o} H_2S\\ H_2S + \dfrac{3}{2}O_2 \xrightarrow{t^o} SO_2 + H_2O\\ 2SO_2 +O_2 \xrightarrow{t^o,xt} 2SO_3\\ SO_3 + H_2O \to H_2SO_4\\ CuO + H_2SO_4 \to CuSO_4 + H_2O\\ CuSO_4 + BaCl_2 \to BaSO_4 + CuCl)2\)
1)
4FeS2 + 11O2 -to-> 2Fe2O3 + 8SO2
SO2 + 2H2S -to-> 3S + 2H2O
S + H2 -to-> H2S
H2S + 4Br2 + 4H2O => H2SO4 + 8HBr
BaCl2 + H2SO4 => BaSO4 + 2HCl
2HCl-dp-> H2 + Cl2
Na + 1/2Cl2 -to-> NaCl
NaCl + AgNO3 => AgCl + NaNO3
2)
H2 + S -to-> H2S
2H2S + 3O2 -to-> 2SO2 + 2H2O
SO2 + 1/2O2 -to,V2O5-> SO3
SO3 + H2O => H2SO4
H2SO4 + CuO => CuSO4 + H2O
CuSO4 + BaCl2 => BaSO4 + CuCl2
3)
MnO2 + 4HCl(đ) -to-> MnCl2 + Cl2 + 2H2O
Cl2 + H2 -as-> 2HCl
FeS + 2HCl => FeCl2 + H2S
2H2S + 3O2 -to-> 2SO2 + 2H2O
2NaOH + SO2 => Na2SO3 + H2O
Na2SO3 + H2SO4 => Na2SO4 + SO2 + H2O
4)
2KMnO4 + 16HCl(đ) => 2KCl + 2MnCl2 + 5Cl2 +8H2O
Cl2 + H2 -as-> 2HCl
Fe(OH)3 + 3HCl => FeCl3 + 3H2O
FeCl3 + 3AgNO3 => Fe(NO3)3 + 3AgCl
AgCl -to-> Ag + 1/2Cl2
2NaBr + Cl2 => 2NaCl + Br2
2NaI + Br2 => 2NaBr + I2