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

b, \(n_{NaOH}=\dfrac{10}{40}=0,25\left(mol\right)\)

Theo PT: \(n_{CuCl_2}=n_{Cu\left(OH\right)_2}=\dfrac{1}{2}n_{NaOH}=0,125\left(mol\right)\)

\(\Rightarrow m_{Cu\left(OH\right)_2}=0,125.98=12,25\left(g\right)\)

c, \(C_{M_{CuCl_2}}=\dfrac{0,125}{0,1}=1,25\left(M\right)\)

\(n_{NaOH}=\dfrac{10}{40}=0,25\left(mol\right)\)

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

                   0,125<-----0,25---------------------->0,125

b) \(m_{Kt}=m_{Cu\left(OH\right)2}=0,125.98=12,25\left(g\right)\)

c) \(C_{MCuCl2}=\dfrac{0,125}{0,1}=1,25\left(M\right)\)

\(n_{NaOH}=\dfrac{10}{40}=0,25\left(mol\right)\)

PTHH:

\(CuCl_2+2NaOH\rightarrow Cu\left(OH\right)_2+2NaCl\)

0,125       0,25                0,125         0,25

\(m_{Cu\left(OH\right)_2}=0,125.98=12,25\left(g\right)\)

\(C_{M\left(CuCl_2\right)}=\dfrac{0,125}{0,1}=1,25\left(M\right)\)

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

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

a. PTHH: \(CuSO_4+2NaOH--->Na_2SO_4+Cu\left(OH\right)_2\downarrow\)

b. Đổi 100ml = 0,1 lít

Ta có: \(n_{Cu\left(OH\right)_2}=\dfrac{9,8}{98}=0,1\left(mol\right)\)

Theo PT: \(n_{CuSO_4}=n_{Cu\left(OH\right)_2}=0,1\left(mol\right)\)

=> \(m_{CuSO_4}=0,1.160=16\left(g\right)\)

c. Theo PT: \(n_{NaOH}=2.n_{CuSO_4}=2.0,1=0,2\left(mol\right)\)

=> \(C_{M_{NaOH}}=\dfrac{0,2}{0,1}=2M\)

PTHH: \(CuCl_2+2NaOH\rightarrow2NaCl+Cu\left(OH\right)_2\downarrow\)

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

Ta có: \(\left\{{}\begin{matrix}n_{CuCl_2}=0,2\cdot2=0,4\left(mol\right)\\n_{NaOH}=0,2\cdot2=0,4\left(mol\right)\end{matrix}\right.\)

Xét tỉ lệ: \(\dfrac{0,4}{1}>\dfrac{0,4}{2}\) \(\Rightarrow\) CuCl₂ còn dư

\(\Rightarrow\left\{{}\begin{matrix}n_{NaCl}=0,4\left(mol\right)\\n_{Cu\left(OH\right)_2}=0,2\left(mol\right)=n_{CuO}=n_{CuCl_2\left(dư\right)}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{CuO}=0,2\cdot80=16\left(g\right)\\C_{M_{NaCl}}=\dfrac{0,4}{0,2+0,2}=1\left(M\right)\\C_{M_{CuCl_2}}=\dfrac{0,2}{0,4}=0,5\left(M\right)\\\end{matrix}\right.\)

a/ PTHH: CO₂ + Ca(OH)₂ ===> CaCO₃ + H₂O

b/ n_CO2 = 2,24 / 22,4 = 0,1 mol

=> n_Ca(OH)2 = n_CO2 = 0,1 mol

=>C_M[Ca(OH)2] = 0,1 / 0,2 = 0,5M

c/ n_CaCO3 = n_CO2 = 0,1 mol

=> m_CaCO3 = 0,1 x 100 = 10 gam

nco2= 2,24/22,4=0,1 mol

co2+ ca(oh)2-> caco3+ h2o

nca(oh)2= nco2=0,1

=> cm ca(oh)2= 0,1/0,2=0,5M

ncaco3= nco2= 0,1

=>m kết tủa=0,1*100=10g

a,2AgNO3+CaCl2=>Ca(NO3)2+2AgCl

b, nAgNO3=17/170=0,1(mol)

Theo PTPU, ta có nAgCl=nAgNO3=0,1(mol)

=>mAgCl=0,1.143,5=14,35(g)

c, ta có 2nCaCl2=nAgNO3=0,05(mol)

=>CM(CaCl2)=0,05/0,1=0,5(M)

a, PT: \(Mg+2HCl\rightarrow MgCl_2+H_2\)

\(MgCl_2+2NaOH\rightarrow2NaCl+Mg\left(OH\right)_{2\downarrow}\)

\(Mg\left(OH\right)_2\underrightarrow{t^o}MgO+H_2O\)

b, Ta có: \(n_{Mg}=\dfrac{9,6}{24}=0,4\left(mol\right)\)

Theo PT: \(n_{HCl}=2n_{Mg}=0,8\left(mol\right)\)

\(\Rightarrow C_{M_{HCl}}=\dfrac{0,8}{0,2}=4\left(M\right)\)

c, Theo PT: \(n_{MgO}=n_{Mg}=0,4\left(mol\right)\)

\(\Rightarrow m_{MgO}=0,4.40=16\left(g\right)\)

\(a,PTHH:CuCl_2+2KOH\rightarrow Cu\left(OH\right)_2\downarrow+2KCl\\ n_{CuCl_2}=\dfrac{27}{135}=0,2\left(mol\right)\\ \Rightarrow n_{KOH}=2n_{CuCl_2}=0,4\left(mol\right)\\ \Rightarrow C\%_{KOH}=\dfrac{0,4}{0,2}=2M\\ b,PTHH:Cu\left(OH\right)_2\rightarrow^{t^o}CuO+H_2O\\ \Rightarrow n_{CuO}=n_{Cu\left(OH\right)_2}=n_{CuCl_2}=0,2\left(mol\right)\\ \Rightarrow m_{CuO}=0,2\cdot80=16\left(g\right)\)