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a)
\(2NaOH+SO_2\rightarrow Na_2SO_3+H_2O\)
\(2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\)
\(2NaOH+Al_2O_3\rightarrow2NaAlO_2+H_2O\)
\(2NaOH+CuSO_4\rightarrow Na_2SO_4+Cu\left(OH\right)_2\downarrow\)
b)
\(2HCl+CuO\rightarrow CuCl_2+H_2O\)
\(2HCl+Cu\left(OH\right)_2\rightarrow CuCl_2+2H_2O\)
\(6HCl+Al_2O_3\rightarrow2AlCl_3+3H_2O\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)
\(2Al + 3CuSO_4 \to Al_2(SO_4)_3 + 3Cu\\ Fe + CuSO_4 \to FeSO_4 + Cu\)
Chất kết tủa : Cu
3 muối tan trong dung dịch : \(Al_2(SO_4)_3,FeSO_4,CuSO_4\) dư.
\(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\\
n_{H_2SO_4}=\dfrac{24,5}{98}=0,25\left(mol\right)\\
pthh:Fe+H_2SO_4\rightarrow FeSO_4+H_2\uparrow\)
\(LTL:\dfrac{0,2}{1}< \dfrac{0,25}{1}\)
=> H2SO4 dư
\(n_{H_2}=n_{H_2SO_4\left(p\text{ư}\right)}=n_{Fe}=0,2\left(mol\right)\\
V_{H_2}=0,2.22,4=4,48l\\
m_{H_2SO_4\left(d\right)}=\left(0,25-0,2\right).98=4,9g\)
a, tác dụng với H2So4:
- mg(OH)2 + h2so4 -> mgso4 + 2h2o
- bacl2 + h2so4 -> baso4 + 2hcl
- caco3 + h2so4 -> caso4 + h2o + co2
- 2kclo3 + h2so4 -> k2so4 + 2hclo3
b, tác dụng với naoh:
mg(no3)2 + 2naoh -> mg(OH)2 + nano3
kclo3+naoh -> koh + naclo3
Câu 1:
a) \(Mg\left(OH\right)_2+H_2SO_4\rightarrow MgSO_4+2H_2O\)
\(CaCO_3+H_2SO_4\rightarrow CaSO_4+CO_2\uparrow+H_2O\)
\(BaCl_2+H_2SO_4\rightarrow BaSO_4\downarrow+2HCl\)
b) \(Mg\left(NO_3\right)_2+2NăOH\rightarrow2NaNO_3+Mg\left(OH\right)_2\downarrow\)
c) \(Mg\left(OH\right)_2\underrightarrow{t^o}MgO+H_2O\)
\(2Mg\left(NO_3\right)_2\underrightarrow{t^o}2MgO+4NO_2\uparrow+O_2\uparrow\)
\(CaCO_3\underrightarrow{t^o}CaO+CO_2\uparrow\)
\(KClO_2\underrightarrow{t^o}KCl+O_2\uparrow\)
a)b)c)d) mBaCl2=150.16,64%=24,96g
=>nBaCl2=0,12 mol
mH2SO4=100.14,7%=14,7g=>nH2SO4=0,15mol
BaCl2 + H2SO4 =>BaSO4 +2HCl
Bđ: 0,12 mol; 0,15 mol
Pứ: 0,12 mol=>0,12 mol=>0,12 mol=>0,24 mol
Dư: 0,03 mol
Dd ban đầu chứa BaCl2 0,12 mol và H2SO4 0,15 mol
Dd A sau phản ứng chứa HCl 0,24 mol và H2SO4 dư 0,03 mol
mHCl=0,24.36,5=8,76g
mH2SO4=0,03.98=2,94g
Kết tủa B là BaSO4 0,12 mol=>mBaSO4=0,12.233=27,96g
mddA=mddBaCl2+mddH2SO4-mBaSO4
=150+100-27,96=222,04g
C%dd HCl=8,76/222,04.100%=3,945%
C% dd H2SO4=2,94/222,04.100%=1,324%
e) HCl +NaOH =>NaCl +H2O
0,24 mol=>0,24 mol
H2SO4 +2NaOH =>Na2SO4 + 2H2O
0,03 mol=>0,06 mol
TÔNG nNaOH=0,3 mol
=>V dd NaOH=0,3/2=0,15 lit
Gọi a, b, c, d là mol mỗi chất trong 32g X
Bảo toàn e: (1)
Bảo toàn e: (2)
Lấy (2) trừ (1) =>
Gọi a, b, c, d là mol mỗi chất trong 32g X
Bảo toàn e: (1)
Bảo toàn e: (2)
Lấy (2) trừ (1) =>
a) \(Mg+H_2SO_4\rightarrow MgSO_4+H_2\)
b) \(n_{Mg}=\dfrac{2,4}{24}=0,1\left(mol\right)\)
Theo PTHH: \(n_{H_2}=n_{Mg}=0,1\left(mol\right)\)
\(\Rightarrow V_{H_2}=0,1.22,4=2,24\left(l\right)\)
c) \(C\%_{MgSO_4}=\dfrac{0,1.120}{2,4+100-0,1.2}.100\%\approx11,74\%\)
\(n_{Zn}=\dfrac{13}{65}=0.2\left(mol\right)\)
\(n_{HCl}=\dfrac{182.5\cdot10}{100\cdot36.5}=0.5\left(mol\right)\)
\(Zn+2HCl\rightarrow ZnCl_2+H_2\)
\(0.2......0.4..........0.2........0.2\)
\(n_{HCl\left(dư\right)}=0.5-0.4=0.1\left(mol\right)\)
\(m_{HCl\left(dư\right)}=0.1\cdot36.5=3.65\left(g\right)\)
\(V_{H_2}=0.2\cdot22.4=4.48\left(l\right)\)
\(m_{\text{dung dịch sau phản ứng}}=13+182.5-0.2\cdot2=195.1\left(g\right)\)
\(C\%_{HCl\left(dư\right)}=\dfrac{3.65}{195.1}\cdot100\%=1.87\%\)
\(C\%_{ZnCl_2}=\dfrac{0.2\cdot136}{195.1}\cdot100\%=13.94\%\)
\(\text{1)}m_{KOH}=40.35\%=14\left(g\right)\\ \rightarrow n_{KOH}=\dfrac{14}{56}=0,25\left(mol\right)\\ PTHH:KOH+HCl\rightarrow KCl+H_2O\\ \text{Theo pthh}:n_{HCl}=n_{KOH}=0,25\left(mol\right)\\ \rightarrow V_{ddHCl}=0,25.0,5=0,125\left(l\right)\)
\(\text{2)}n_{Al}=\dfrac{4,05}{27}=0,15\left(mol\right)\\ n_{H_2SO_4}=200.14,7\%=29,4\left(g\right)\\ \rightarrow n_{H_2SO_4}=\dfrac{29,4}{98}=0,3\\ \text{PTHH}:2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\uparrow\\ \text{LTL}:\dfrac{0,15}{2}< \dfrac{0,3}{3}\rightarrow H_2SO_4\text{ dư}\)
\(\text{Theo pthh}:\left\{{}\begin{matrix}n_{H_2SO_4\left(pư\right)}=\dfrac{3}{2}n_{Al}=\dfrac{3}{2}.0,15=0,225\left(mol\right)\\n_{H_2}=n_{H_2SO_4\left(pư\right)}=0,225\left(mol\right)\\n_{Al_2\left(SO_4\right)_3}=\dfrac{1}{2}n_{Al}=\dfrac{1}{2}.0,15=0,075\left(mol\right)\end{matrix}\right.\\ \rightarrow m_{dd\left(\text{sau phản ứng}\right)}=200+4,05-0,3.2=203,45\left(g\right)\)
\(\rightarrow\left\{{}\begin{matrix}C\%_{H_2SO_4\text{ dư}}=\dfrac{\left(0,3-0,225\right).98}{203,45}=3,61\%\\C\%_{Al_2\left(SO_4\right)_3}=\dfrac{342.0,075}{203,45}=12,61\%\end{matrix}\right.\)
2NaOH + CuSO4 → Cu(OH)2 + Na2SO4
n NaOH = 0,2.5 = 1(mol)
n CuSO4 = 0,1.2 = 0,2(mol)
Ta có :
n NaOH / 2 = 0,5 > n CuSO4 / 1 = 0,2 => NaOH dư
n Cu(OH)2 = n CuSO4 = 0,2 mol
=> m A = 0,2.98 = 19,6 gam
n Na2SO4 = n CuSO4 = 0,2 mol
n NaOH pư = 2n CuSO4 = 0,4(mol)
V dd = 0,2 + 0,1 = 0,3(lít)
Suy ra:
CM Na2SO4 = 0,2/0,3 = 0,67M
CM NaOH = (1 - 0,4)/0,3 = 2M