a, \(CuCl_2+2NaOH\rightarrow2NaCl+Cu\left(OH\right)_2\)

\(Cu\left(OH\right)_2\underrightarrow{t^o}CuO+H_2O\)

b, \(n_{CuCl_2}=\dfrac{27}{135}=0,2\left(mol\right)\)

Theo PT: \(n_{CuO}=n_{Cu\left(OH\right)_2}=n_{CuCl_2}=0,2\left(mol\right)\)

\(\Rightarrow m_{CuO}=0,2.80=16\left(g\right)\)

c, \(n_{NaOH}=2n_{CuCl_2}=0,4\left(mol\right)\)

\(\Rightarrow C\%_{NaOH}=\dfrac{0,4.40}{200}.100\%=8\%\)

\(a,PTHH:3NaOH+FeCl_3\rightarrow3NaCl+Fe\left(OH\right)_3\downarrow\\ 2Fe\left(OH\right)_3\rightarrow^{t^o}Fe_2O_3+3H_2O\uparrow\\ b,n_{FeCl_3}=1,5\cdot0,2=0,3\left(mol\right)\\ \Rightarrow n_{NaOH}=3n_{FeCl_3}=0,9\left(mol\right)\\ \Rightarrow V_{dd_{NaOH}}=\dfrac{0,9}{2}=0,45\left(l\right)\)

Theo đề: \(\left\{{}\begin{matrix}X:Fe\left(OH\right)_3\\A:NaCl\\Y:Fe_2O_3\end{matrix}\right.\)

Theo PT: \(n_{NaCl}=3n_{FeCl_3}=0,9\left(mol\right)\)

\(\Rightarrow C_{M_{NaCl}}=\dfrac{0,9}{0,45+0,2}\approx1,4M\)

\(c,\) Theo PT: \(n_{Fe\left(OH\right)_3}=n_{FeCl_3}=0,3\left(mol\right);n_{Fe_2O_3}=\dfrac{1}{2}n_{Fe\left(OH\right)_3}=0,15\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}m_X=m_{Fe\left(OH\right)_3}=0,3\cdot107=32,1\left(g\right)\\m_Y=m_{Fe_2O_3}=0,15\cdot160=24\left(g\right)\end{matrix}\right.\)

PTHH: \(HNO_3+NaOH\rightarrow NaNO_3+H_2O\)

Ta có: \(n_{HNO_3}=0,2\cdot0,15=0,03\left(mol\right)=n_{NaNO_3}\)

\(\Rightarrow C_{M_{NaNO_3}}=\dfrac{0,03}{0,2+0,3}=0,06\left(M\right)\)

\(\text{nMgCl2 = 0,2.0,15 = 0,03 mol}\)

\(\text{a) (1) MgCl2 + 2NaOH → Mg(OH)2 ↓ + 2NaCl}\)

\(\text{ (2) Mg(OH)2 --to--> MgO + H2O}\)

b)

Từ các PTHH: nMgO(2) = nMg(OH)2 (2) = nMg(OH)2 (1) = nMgCl2 = 0,03 mol

\(\text{→ m = mMgO = 0,03.40 = 1,2 (g)}\)

c) Theo PTHH (1): nNaCl = 2nMgCl2 = 2.0,03 = 0,06 mol

\(\text{→ CM NaCl = n : V = 0,06 : (0,2 + 0,3) = 0,12M}\)

a) MgCl2+2NaOH---->Mg(OH)2+2NaCl

Mg(OH)2--->MgO+H2O

b) n MgCl2=0,2.0,15=0,03(mol)

Theo pthh1

n Mg(OH)2=n MgCl2=0,03(mol0

Theo pthh2

n MgO=n Mg(OH)2=0,03(mol)

m MgO=0,03.40=1,2(g)

c) V dd =0,2+0,3=0,5(l)

Theo pthh2

n NaCl=2nMgCl2=0,06(mol)

CM NaCl=\(\frac{0,06}{0,5}=0,12\left(M\right)\)

\(n_{CuCl_2}=\dfrac{2,7}{135}=0,02\left(mol\right)\\ a,CuCl_2+2NaOH\rightarrow Cu\left(OH\right)_2+2NaCl\\ b,n_{Cu\left(OH\right)_2}=n_{CuCl_2}=0,02\left(mol\right)\\ n_{NaCl}=n_{NaOH}=2.0,02=0,04\left(mol\right)\\ b,m_D=m_{Cu\left(OH\right)_2}=98.0,02=1,96\left(g\right)\\ Cu\left(OH\right)_2\rightarrow\left(t^o\right)CuO+H_2O\\ n_{CuO}=n_{Cu\left(OH\right)_2}=0,02\left(mol\right)\\ \Rightarrow m_E=m_{CuO}=0,02.80=1,6\left(g\right)\)