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câu 1
a) CuO+H2SO4-->CuSO4 + H2O
\(m_{h2so4}=196.5\%=9.6\left(g\right)\)
\(n_{h2so4}=\dfrac{9.8}{98}=0.1\left(mol\right)\)
\(n_{CuO}=n_{h2SO4}=0.1\left(mol\right)\)
\(m_{CuO}=80\cdot0,1=8\left(g\right)\)
b)
\(m_{CuSO4}=160\cdot0,1=16\left(g\right)\)
\(m_{ddCuSO4}=8+196=204\left(g\right)\)
\(C\%_{MgSO4}=\dfrac{16}{204}\cdot100\%=7,84\%\)
c)\(n_{MgO}=n_{H2SO4}=0.1\left(mol\right)\)
\(m_{MgO}=n\cdot M=0.1\cdot40=4\left(g\right)\)
a, \(CuO+2HCl\rightarrow CuCl_2+H_2O\)
b, \(n_{HCl}=0,2.0,5=0,1\left(mol\right)\)
Theo PT: \(n_{CuO}=n_{CuCl_2}=\dfrac{1}{2}n_{HCl}=0,05\left(mol\right)\)
\(\Rightarrow m_{CuO}=0,05.80=4\left(g\right)\)
c, \(C_{M_{CuCl_2}}=\dfrac{0,05}{0,2}=0,25\left(M\right)\)
\(n_{HCl}=0,2.0,5=0,1\left(mol\right)\)
PTHH :
\(CuO+2HCl\rightarrow CuCl_2+H_2O\)
0,05 0,1 0,05
\(b,m_{CuO}=0,05.80=4\left(g\right)\)
\(c,C_{M\left(CuCl_2\right)}=\dfrac{0,05}{0,2}=0,25\left(M\right)\)
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 = \(\frac{5,85}{53+18,25-2,2}\approx8,47\%\)
PTHH: MgO + 2 HCl -> MgCl2 + H2O
- Muối thu dc sau phản ứng là MgCl2.
nMgCl2= 19/95= 0,2(mol)
a) nMgO= nMgCl2= 0,2(mol)
=> mMgO= 0,2.40= 8(g)
b) nHCl= 2.0,2= 0,4(mol)
=> mHCl= 0,4.36,5= 14,6(g)
=> mddHCl = (14,6.100)/10= 146(g)
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
\(m_{FeCl_3}=\dfrac{1}{6}.162,5\approx27g\)
\(C_{MddFeCl_3}=\dfrac{1}{6}:0,5\approx0,3M\)
Làm nhanh zùm mk.
thank all.
Ai ơi làm nhanh zùm mk nha
mk cần gấp lắm rồi
huhu.