a)n_CuO= 16/80=0.2(mol)

CuO + H₂SO₄ ➞ CuSO₄ +H₂O

0.2........0.2.............0.2.........0.2...(mol)

m_CuSO4=0.2*160=32(g)

m_H2SO4=0.2*98=19.6(g)

m_{dung dịch H2SO4}=19.6*100/10=196(g)

m_{dung dịch mới thu được}=196+16=212(g)

C%_CuSO4=32/212*100%=15.094%

b) CuSO₄ + 2 NaOH ➞ Cu(OH)₂ + Na₂SO₄

......0.2...............0.4.............0.2..............0.2.......(mol)

V_NaOH= 0.4/2=0.2(l)

m_Cu(OH)2=0.2*98=19.6(g)

hình như không có phản ứng của axit sunfurơ với CuO bạn nhé

nCuO = \(\frac{16}{80}\)= 0,2 mol

PTHH: CuO + H2SO4 → CuSO4 + H2O

_______0,2____ 0,2______0,2___________ (mol)

→ mH2SO4 = 0,2.98 = 19,6 mol

→ m dd H2SO4 = 19,6.(100/9,8) = 200 gam

BTKL → m dd sau pư = mCuO + m dd H2SO4 = 16 + 200 = 216 gam

\(a,C\%_{CuSO4}=\frac{0,2.160}{216}.100\%=14,8\%\)

b) CuSO4 + 2NaOH → Cu(OH)2 + Na2SO4

___ 0,2______0,4______ 0,2_____________

→ V dd NaOH = 0,4/2 = 0,2 lít = 200 ml

→ m kết tủa = mCu(OH)2 = 0,2.98 = 19,6 gam

a)nCuO= 16/80=0.2(mol)

CuO + H2SO4 ➞ CuSO4 +H2O

0.2........0.2.............0.2.........0.2...(mol)

mCuSO4=0.2*160=32(g)

mH2SO4=0.2*98=19.6(g)

mdung dịch H2SO4=19.6*100/10=196(g)

mdung dịch mới thu được=196+16=212(g)

C%CuSO4=32/212*100%=15.094%

b) CuSO4 + 2 NaOH ➞ Cu(OH)2 + Na2SO4

......0.2...............0.4.............0.2..............0.2.......(mol)

VNaOH= 0.4/2=0.2(l)

mCu(OH)2=0.2*98=19.6(g)

PTHH: \(KOH+HCl\rightarrow KCl+H_2O\)

            \(KCl+AgNO_3\rightarrow KNO_3+AgCl\downarrow\)

Ta có: \(n_{HCl}=0,2\cdot2=0,4\left(mol\right)=n_{KOH}=n_{KNO_3}=n_{AgCl}\)

\(\Rightarrow\left\{{}\begin{matrix}m_{KOH}=0,1\cdot56=5,6\left(g\right)\\m_{AgCl}=0,1\cdot143,5=14,35\left(g\right)\\m_{KNO_3}=0,1\cdot101=10,1\left(g\right)\end{matrix}\right.\)

cho cac axit :HCLO,HNO3,H2S,H2SO3,HNO2,HCLO4,HMno4.so axit manh la

Rainbow

BaO+H2O -> Ba(OH)2 
0,02             0,02  
a) CM = n/V = 0,02/0,02 = 1M
b) Ba(OH)2 + H2SO4 -> BaSO4 +2H2O
      0,02          0,02
=> m = 0,392 g 
D = m/V = 1,14
=> 0,392/V = 1,14 => V = 0,34l

a) $CuSO_4 + 2NaOH \to Cu(OH)_2 + Na_2SO_4$

b) $n_{Cu(OH)_2} = n_{CuSO_4} = \dfrac{16}{160} = 0,1(mol)$
$m_{Cu(OH)_2} = 0,1.98 = 9,8(gam)$

c) $n_{NaOH} = 2n_{CuSO_4} = 0,2(mol) \Rightarrow C_{M_{NaOH}} = \dfrac{0,2}{0,1} = 2M$

PTHH: \(CuSO_4+2NaOH\rightarrow Na_2SO_4+Cu\left(OH\right)_2\downarrow\)

Ta có: \(n_{CuSO_4}=\dfrac{16}{160}=0,1\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}n_{NaOH}=0,2\left(mol\right)\\n_{Cu\left(OH\right)_2}=0,1\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}C_{M_{NaOH}}=\dfrac{0,2}{0,1}=2\left(M\right)\\m_{Cu\left(OH\right)_2}=0,1\cdot98=9,8\left(g\right)\end{matrix}\right.\)