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Bài 1: \(n_{H_2SO_4}=\frac{9}{49}\left(mol\right)\)
H2SO4 + 2KOH -> K2SO4 + 2H2O
=> nKOH= 2nH2SO4 = \(\frac{18}{49}\left(mol\right)\)
=> Vdd KOH = \(\frac{18}{49}:\frac{2}{1000}=\frac{9000}{49}\left(ml\right)\)
b) nK2SO4 = nH2SO4 = \(\frac{9}{49}\left(mol\right)\)
=> mK2SO4= \(\frac{9}{49}\cdot174=\frac{1566}{49}\left(g\right)\)
mdd KOH = \(\frac{9000}{49}\cdot1,12=\frac{1440}{7}\left(g\right)\)
c) \(\%m_{K_2SO_4}=\frac{1566}{49}:\left(200+\frac{1440}{7}\right)\cdot100\%\approx7,87\%\)
bài 2: nNa2CO3 = 0,05 (mol)
PTHH:
Na2CO3 + 2HCl -> 2NaCl + H2O + CO2
=> nHCl = n NaCl = 2nNa2CO3 = 0,1 (mol)
=> mNaCl= 0,1 . 58,5 = 5,85 (g)
b) nCO2 = nNa2CO3 = 0,05 (mol)
=> mCO2 = 0,05 . 44 = 2,2 (g)
mdd HCl = 0,1 . 36,5 :20% = 18,25 (g)
=> %mNaCl = \(\frac{5,85}{53+18,25-2,2}\approx8,47\%\)
Bài 1: \(Ca\left(OH\right)_2\left(0,3\right)+2HCl\left(0,6\right)\rightarrow2CaCl_2\left(0,6\right)+2H_2O\)
\(n_{Ca\left(OH\right)_2}=0,3\left(mol\right)\)
\(\Rightarrow V_{ddHCl}=\dfrac{0,6}{0,2}=3M\)
\(m_{CaCl_2}=0,6.111=66,6\left(g\right)\)
\(C_{MddCaCl_2}=\dfrac{0,6}{0,3}=2M.\)
Bài 2: \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)
\(n_{Fe}=0,1\left(mol\right)\)
\(n_{H_2SO_4}=0,1\left(mol\right)\)
=> Pư này pư vừa đủ
\(V_{H_2}=0,1.22,4=2,24\left(l\right)\)
\(C_{MddFeSO_4}=\dfrac{0,1}{0,5}=0,2M\)
\(n_{FeSO_4}=n_{FeSO_4.7H_2O}=0,1\left(mol\right)\)
\(\Rightarrow m_{FeSO_4.7H_2O}=0,1.278=27,8\left(g\right)\)
\(4H_2\left(0,1\right)+Fe_3O_4\left(0,025\right)\rightarrow3Fe+4H_2O\)
\(\Rightarrow m_{Fe_3O_4}=0,025.232=5,8\left(g\right).\)
1)
a dd KOH
MgCl2 + 2KOH --------> Mg(OH)2 + 2KCl
Cu(NO3)2 + 2KOH ------> Cu(OH)2 + 2KNO3
b) AgNO3
2AgNO3 + MgCl2 -------> 2AgCl + Mg(NO3)2
nNa2O=15,5/62=0,25mol
pt : Na2O + H2O ---------> 2NaOH
npứ: 0,25---------------------->0,5
CM(NaOH)=0,5/0,5=1M
pt : 2NaOH + H2SO4 ------> Na2SO4 + 2H2O
npứ:0,5---------->0,25
mH2SO4 = 0,25.98=24,5g
mddH2SO4 =\(\dfrac{24,5.100}{20}=122,5\)
Vdd H2SO4=122,5/1,14\(\approx107,46ml\)
Gọi nNa2CO3 = x (mol)
Na2CO3 + 2HCl \(\rightarrow\) 2NaCl + H2O + CO2
x \(\rightarrow\) 2x \(\rightarrow\) 2 x (mol)
C%(NaCl) = \(\frac{2.58,5x}{200+120}\) . 100% = 20%
=> x =0,547 (mol)
mNa2CO3 = 0,547 . 106 = 57,982 (g)
mHCl = 2 . 0,547 . 36,5 =39,931 (g)
C%(Na2CO3) =\(\frac{57,892}{200}\) . 100% = 28,946%
C%(HCl) = \(\frac{39,931}{120}\) . 100% = 33,28%
Bài 1:
PTHH: \(BaO+H_2SO_4\rightarrow BaSO_4+H_2O\)
Bđ____0,05___0,2
Pư____0,05___0,05_______0,05
Kt____0______0,15_______0,05
\(m_{kt}=m_{BaSO_4}=0,05.233=11,65\left(g\right)\)
\(m_{ddsaupư}=7,65+200-11,65=196\left(g\right)\)
\(C\%ddH_2SO_4=7,5\%\)
Bài 2: \(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\)
bđ___0,1_______0,5
pư__1/12_______0,5_____1/6
kt ___1/60______0_______1/6
\(m_{FeCl_3}=\dfrac{1}{6}.162,5\approx27g\)
\(C_{MddFeCl_3}=\dfrac{1}{6}:0,5\approx0,3M\)
ZnO + H2SO4 = ZnSO4 + H2O
0.1mol:2.32mol
=> H2SO4 dư theo ZnO
=> khối lượng axits tham gia: 0,1.(2+32+16.4)=9.8g
=> khối lượng muối : mZnSO4=0.1(65+32+16.4)=16.1g
nồng độ mol sau pu: CM=\(\frac{0.1}{0.58}\)=\(\frac{5}{29}\)
Làm nhanh zùm mk