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- HCl:
\(AgNO_3+HCl\rightarrow AgCl\downarrow+HNO_3\\ Na_2CO_3+2HCl\rightarrow2NaCl+CO_2\uparrow+H_2O\\ Ba\left(NO_3\right)_2+Na_2CO_3\rightarrow BaCO_3\downarrow+2NaNO_3\\ KHSO_3+HCl\rightarrow KCl+SO_2\uparrow+H_2O\\ MgCO_3+2HCl\rightarrow MgCl_2+CO_2\uparrow+H_2O\\ Ba\left(OH\right)_2+2HCl\rightarrow BaCl_2+2H_2O\\ K_2SO_3+2HCl\rightarrow2KCl+CO_2\uparrow+H_2O\\ 2HCl+Ba\left(HCO_3\right)_2\rightarrow BaCl_2+2H_2O+2CO_2\uparrow\)
- H2SO4:
\(2AgNO_3+H_2SO_4\rightarrow Ag_2SO_4\downarrow+2HNO_3\\ Na_2CO_3+H_2SO_4\rightarrow Na_2SO_4+CO_2\uparrow+H_2O\\ Ba\left(NO_3\right)_2+H_2SO_4\rightarrow BaSO_4\downarrow+2HNO_3\\ 2KCl+H_2SO_4\rightarrow K_2SO_4+2HCl\uparrow\\ CaSO_3+H_2SO_4\rightarrow CaSO_4\downarrow+SO_2\uparrow+H_2O\\ BaCl_2+H_2SO_4\rightarrow BaSO_4\downarrow+2HCl\\ Pb\left(NO_3\right)_2+H_2SO_4\rightarrow PbSO_4\downarrow+2HNO_3\)
\(H_2SO_4+MgCO_3\rightarrow MgSO_4+CO_2\uparrow+H_2O\\ H_2SO_4+Ba\left(OH\right)_2\rightarrow BaSO_4\downarrow+2H_2O\\ K_2SO_3+H_2SO_4\rightarrow K_2SO_4+SO_2\uparrow+H_2O\)
\($a/$\\ Zn+2HCl \to ZnCl_2+H_2\\ b/\\ n_{Zn}=0,1(mol)\\ n_{HCl}=0,1.2=0,2(mol)\\ m_{ddHCl}=\frac{0,2.36,5.100}{7,3}=100(g)\\ C\%_{ZnCl_2}=\frac{0,1.136}{100+6,5-0,1.2}.100\%=12,8\%\)
\(a/\\ Zn+2HCl \to ZnCl_2+H_2\\ b/\\ n_{Zn}=\frac{6,5}{65}=0,1(mol)\\ n_{HCl}=0,2(mol)\\ m_{ddsaupu}=\frac{0,1.36,5.100}{7,3}+6,5-0,1.2=56,3(g)\\ C\%_{ZnCl_2}=\frac{0,1.136}{56,3}.100\%=24,16\%\)
- Những kim loại tác dụng được với dd H2SO4 loãng: Al, Fe.
\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\\ Fe+H_2SO_4\rightarrow FeSO_4+H_2\)
- Những kim loại tác dụng được với dung dịch AgNO3 là Al, Fe, Cu.
\(Al+3AgNO_3\rightarrow Al\left(NO_3\right)_3+3Ag\\ Fe+2AgNO_3\rightarrow Fe\left(NO_3\right)_2+2Ag\\ Cu+2AgNO_3\rightarrow Cu\left(NO_3\right)_2+2Ag\)
- Kim loại tác dụng được vớ dung dịch NaOH: Al.
\(NaOH+Al+H_2O\rightarrow NaAlO_2+\dfrac{3}{2}H_2\)
- Những kim loại tác dụng với dung dịch H2SO4 đặc, nguội: Cu, Ag.
\(Cu+2H_2SO_{4\left(đặc.nguội\right)}\rightarrow CuSO_4+SO_2+2H_2O\\ 2Ag+2H_2SO_{4\left(đặc.nguội\right)}\rightarrow Ag_2SO_4+SO_2+2H_2O\)
- Những kim loại tác dụng với dung dịch H2SO4 đặc, nóng: Al, Fe, Cu, Ag
\(2Al+6H_2SO_{4\left(đặc.nóng\right)}\rightarrow Al_2\left(SO_4\right)_3+3SO_2+6H_2O\\ 2Fe+6H_2SO_{4\left(đặc,nóng\right)}\rightarrow Fe_2\left(SO_4\right)_3+3SO_2+6H_2O\\ Cu+2H_2SO_{4\left(đặc,nóng\right)}\rightarrow CuSO_4+SO_2+2H_2O\\ 2Ag+2H_2SO_{4\left(đặc,nóng\right)}\rightarrow Ag_2SO_4+SO_2+2H_2O\)
Giải:
a) Số mol khí CO2 sinh ra là:
nCO2 = V/22,4 = 4,48/22,4 = 0,2 (mol)
PTHH: Na2CO3 + 2HCl -> 2NaCl + H2CO3
PTHH: 10NaHCO3 + 10HCl -> 10NaCl + H2O + 15CO2↑
--------------\(\dfrac{2}{15}\)------------------------------------------0,2--
b) Khối lượng NaHCO3 là:
mNaHCO3 = n.M = \(\dfrac{2}{15}\).84 = 11,2 (g)
Thành phần phần trăm theo khối lượng của NaHCO3 trong hỗn hợp ban đầu là:
%mNaHCO3 = (mNaHCO3/mhh).100 = (11,2/19).100 ≃ 58,95 %
=> %mNa2CO3 = 100 - 58,95 = 41,05 %
Vậy ...
\(\text{a) }Na_2CO_3+2HCl\rightarrow2NaCl+CO_2+H_2O\left(1\right)\\ NaHCO_3+HCl\rightarrow NaCl+CO_2+H_2O\left(2\right)\)
\(\text{b) }n_{CO_2}=\dfrac{V}{22,4}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
\(Na_2CO_3+2HCl\rightarrow2NaCl+CO_2+H_2O\left(1\right)\\ \text{ }\text{ }\text{ }x\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }x\\ NaHCO_3+HCl\rightarrow NaCl+CO_2+H_2O\left(2\right)\\ \text{ }\text{ }\text{ }y\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }y\)
Từ \(\left(1\right)\) và \(\left(2\right),\) ta có hệ phương trình: \(\left\{{}\begin{matrix}x+y=0,2\\106x+84y=19\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,1\\y=0,1\end{matrix}\right.\)
\(\Rightarrow m_{Na_2CO_3}=n\cdot M=0,1\cdot106=10,6\left(g\right)\\ m_{NaHCO_3}=n\cdot M=0,1\cdot84=8,4\left(g\right)\)
\(\Rightarrow\%Na_2CO_3=\dfrac{10,6\cdot100}{19}=55,79\%\\ \%NaHCO_3=\dfrac{8,4\cdot100}{19}=44,21\%\)
Sửa đề : 200ml thành 200g
a) \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\) (1)
\(3H_2+Fe_2O_3-^{t^o}\rightarrow2Fe+3H_2O\) (2)
b) 1/2 lượng khí B: \(n_{H_2\left(2\right)}=3n_{Fe_2O_3}=3.\dfrac{38,4}{160}=0,72\left(mol\right)\)
=> \(n_{H_2\left(1\right)}=0,72.2=1,44\left(mol\right)\)
\(n_{H_2SO_4}=n_{H_2\left(2\right)}=1,44\left(mol\right)\)
=> \(C\%H_2SO_4=\dfrac{1,44.98}{200}.100=70,56\%\)
\(n_{Al}=\dfrac{2}{3}n_{H_2\left(2\right)}=0,96\left(mol\right)\)
=> \(m_{Al}=0,96.27=25,92\left(g\right)\)
\(n_{H_2}= \dfrac{8,96}{22,4} = 0,4(mol)\\ 2Al + 6HCl \to 2AlCl_3 + 3H_2\\ Fe + 2HCl \to FeCl_2 + H_2\\ m_{tăng} = m_{kim\ loại} - m_{H_2} = 11 - 0,4.2 = 10,2(gam)\)
PTHH: 2Al+6HCl→2AlCl3+3H2↑2Al+6HCl→2AlCl3+3H2↑
Fe+2HCl→FeCl2+H2↑Fe+2HCl→FeCl2+H2↑
Ta có: mH2=8,9622,4⋅2=0,8(g)<mhh=11(g)mH2=8,9622,4⋅2=0,8(g)<mhh=11(g)
⇒⇒ Sau p/ứ dd tăng 11−0,8=10,2(g)
https://hoc24.vn/cau-hoi/cho-400-g-dung-dich-naoh-30-tac-dung-vua-het-voi-100-g-dung-dich-hcl-tinh-a-nong-do-muoi-thu-duoc-sau-phan-ungb-tinh-nong-do-axit-hcl-bie.7974814552205
Tui trả lời rùi nghen
\(H2SO4+Na2CO3\rightarrow Na2SO4+H2O+CO2\)
b) \(n_{H2SO4}=\frac{9,8}{98}=0,1\left(mol\right)\)
\(n_{Na2SO4}=n_{H2SO4}=0,1\left(mol\right)\)
\(m_{Na2SO4}=0,1.142=14,2\left(g\right)\)
c) \(n_{CO2}=n_{H2SO4}=0,1\left(mol\right)\)
\(m_{CO2}=0,1.44=4,4\left(g\right)\)
\(m_{dd}sau-pư=9,8+100-4,4=105,4\left(g\right)\)
\(C\%_{Na2SO4}=\frac{14,2}{105,4}.100\%=13,47\%\)
a, \(Na_2CO_3+2HCl\rightarrow2NaCl+CO_2+H_2O\)
\(CaSO_3+2HCl\rightarrow CaCl_2+SO_2+H_2O\)
\(MgSO_3+2HCl\rightarrow MgCl_2+SO_2+H_2O\)
\(BaCO_3+2HCl\rightarrow BaCl_2+CO_2+H_2O\)
\(KHCO_3+HCl\rightarrow KCl+CO_2+H_2O\)
b, \(Na_2CO_3+H_2SO_4\rightarrow Na_2SO_4+CO_2+H_2O\)
\(CaSO_3+H_2SO_4\rightarrow CaSO_4+SO_2+H_2O\)
\(MgSO_3+H_2SO_4\rightarrow MgSO_4+SO_2+H_2O\)
\(BaCO_3+H_2SO_4\rightarrow BaSO_4+CO_2+H_2O\)
\(2KHCO_3+H_2SO_4\rightarrow K_2SO_4+2CO_2+2H_2O\)