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\(CaCO_3 + 2HCl \to CaCl_2 + CO_2 +H_2O\\ CO_2 + NaOH \to NaHCO_3 2NaHCO_3 \xrightarrow{t^o} Na_2CO_3 + CO_2 + H_2O\\ Na_2CO_3 + BaCl_2 \to BaCO_3 + 2NaCl\)
1)
\(CaCO_3\underrightarrow{t^o}CO_2+CaO\\ NaOH+CO_2\rightarrow NaHCO_3\\ NaHCO_3+NaOH\rightarrow Na_2CO_3+H_2O\\ Na_2CO_3+Ba\left(OH\right)_2\rightarrow NaOH+BaCO_3\)
2)
\(n_{HCl}=C_{M_{HCl}}.V_{HCl}=1.0,2=0,2\left(mol\right)\)
PTHH: \(K_2CO_3+2HCl\rightarrow2KCl+H_2O+CO_2\)
\(\Rightarrow n_{K_2CO_3}=n_{CO_2}=0,1\left(mol\right)\)
a) \(V_{CO_2\left(đktc\right)}=0,1.22,4=2,24\left(l\right)\)
b) \(m_{K_2CO_3}=0,1.138=13,8\left(g\right)\)
\(m_{ddK_2CO_3}=\dfrac{13,8.100}{13,8}=100\left(g\right)\)
1:
a) \(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)
PTHH: \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\uparrow\)
______0,2------>0,2------------------->0,2_____(mol)
=> \(V_{H_2}=0,2.22,4=4,48\left(l\right)\)
b) \(V_{ddH_2SO_4}=\dfrac{0,2}{1}=0,2\left(l\right)\)
2:
a)
\(n_{HCl}=2.0,2=0,4\left(mol\right)\)
PTHH: \(Mg+2HCl\rightarrow MgCl_2+H_2\uparrow\)
______0,2<------0,4------------------>0,2______(mol)
=> \(m_{Mg}=0,2.24=4,8\left(g\right)\)
b) \(V_{H_2}=0,2.22,4=4,48\left(l\right)\)
$n_{H_2SO_4}=\dfrac{300}{1000}.1=0,3(mol)$
$2NaOH+H_2SO_4\to Na_2SO_4+2H_2O$
Theo PT: $n_{NaOH}=2n_{H_2SO_4}=0,6(mol)$
$\to m_{NaOH}=0,6.40=24(g)$
a, Ta có: \(m_{NaOH}=200.4\%=8\left(g\right)\) \(\Rightarrow n_{NaOH}=\dfrac{8}{40}=0,2\left(mol\right)\)
PT: \(2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\)
_____0,2______0,1_______0,1 (mol)
\(\Rightarrow V_{ddH_2SO_4}=\dfrac{0,1}{0,2}=0,5M\)
b, Ta có: m dd sau pư = m dd NaOH + m ddH2SO4 = 200 + 50 = 250 (g)
\(\Rightarrow C\%_{Na_2SO_4}=\dfrac{0,1.142}{250}.100\%=5,68\%\)
Bạn tham khảo nhé!
nCuSO4=0,01 mol
Fe+CuSO4=> FeSO4+Cu
0,01 mol =>0,01 mol
mCu=0,01.64=0,64gam
FeSO4+2NaOH=>Fe(OH)2 +Na2SO4
0,01 mol=>0,02 mol
Vdd NaOH=0,02/1=0,02 lit
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
\(n_{NaOH}=\dfrac{200\cdot4\%}{40}=0.2\left(mol\right)\)
\(2NaOH+H_2SO_4\rightarrow Na_2SO_4+H_2O\)
\(0.2..............0.1..............0.1\)
\(V_{dd_{H_2SO_4}}=\dfrac{0.1}{0.2}=0.5\left(l\right)\)
\(m_{Na_2SO_4}=0.1\cdot142=14.2\left(g\right)\)
\(m_{dd}=200+510=710\left(g\right)\)
\(C\%_{Na_2SO_4}=\dfrac{14.2}{710}\cdot100\%=2\%\)
Ta có: mNaOH = 200.4% = 8 (g)
\(\Rightarrow n_{NaOH}=\dfrac{8}{40}=0,2\left(mol\right)\)
PT: \(2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\)
_____0,2______0,1_______0,1 (mol)
a, \(V_{ddH_2SO_4}=\dfrac{0,1}{0,2}=0,5\left(l\right)\)
b, Chất có trong dd sau pư là Na2SO4.
Ta có: m dd sau pư = m dd NaOH + m dd H2SO4 = 200 + 510 = 710 (g)
\(\Rightarrow C\%_{Na_2SO_4}=\dfrac{0,1.142}{710}.100\%=2\%\)
Bạn tham khảo nhé!
\(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\\ PTHH:Fe+2HCl\rightarrow FeCl_2+H_2\)
0,3 0,6 0,3 0,3
\(FeCl_2+2AgNO_3\rightarrow Fe\left(NO_3\right)_3+2AgCl\)
0,3 0,6
\(\rightarrow\left\{{}\begin{matrix}a=0,3.56=16,8\left(g\right)\\b=0,6.143,5=86,1\left(g\right)\end{matrix}\right.\)
\(m_{ddHCl}=150.1,2=180\left(g\right)\\ m_{HCl}=0,6.36,5=21,9\left(g\right)\\ \rightarrow\left\{{}\begin{matrix}C\%_{HCl}=\dfrac{21,9}{180}=12,17\%\\C_{M\left(HCl\right)}=\dfrac{0,6}{0,15}=4M\end{matrix}\right.\)
Bài 1 :
\(a) CaCO_3 \xrightarrow{t^o} CaO + CO_2\\ 2Fe(OH)_3 \xrightarrow{t^o} Fe_2O_3 + 3H_2O\\ b) CaCO_3 + 2HCl \to CaCl_2 + CO_2 + H_2O\\ 2Fe(OH)_3 + 6HCl \to 2FeCl_3 + 6H_2O\\ NaOH + HCl \to NaCl + H_2O\\ c) 2AgNO_3 + 2NaOH \to 2NaNO_3 + Ag_2O + H_2O\\ NaCl + AgNO_3 \to AgCl + NaNO_3\)
Bài 2 :
\(a)2AgNO_3 + BaCl_2 \to 2AgCl + Ba(NO_3)_2\\ n_{AgCl} = n_{AgNO_3} = 0,2.1 = 0,2(mol)\\ \Rightarrow m_{AgCl} = 0,2.143,5 = 28,7(gam)\\ b) n_{BaCl_2} = \dfrac{1}{2}n_{AgNO_3} = 0,1(mol)\\ V_{dd\ BaCl_2} = \dfrac{0,1}{2} = 0,05(lít)\)