\(a.n_{Mg\left(OH\right)_2}=\dfrac{17,4}{58}=0,3\left(mol\right)\\ Mg\left(OH\right)_2+2HCl\rightarrow MgCl_2+2H_2O\\ n_{HCl}=2n_{Mg\left(OH\right)_2}=0,6\left(mol\right)\\ CM_{HCl}=\dfrac{0,6}{0,2}=3M\\b. n_{Mg\left(OH\right)_2}=n_{MgCl_2}=0,3\left(mol\right)\\ m_{MgCl_2}=0,3.85=25,5\left(g\right)\\c.CM_{MgCl_2}=\dfrac{0,3}{0,2}=1,5M \)

PTHH: \(CaCl_2+Na_2CO_3\rightarrow CaCO_3\downarrow+2NaCl\)

a+b) Ta có: \(n_{CaCl_2}=0,1\cdot2=0,2\left(mol\right)\) 

\(\Rightarrow\left\{{}\begin{matrix}n_{CaCO_3}=0,2\left(mol\right)\\n_{NaCl}=0,4\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{CaCO_3}=0,2\cdot100=20\left(g\right)\\C_{M_{NaCl}}=\dfrac{0,4}{0,1+0,2}\approx1,33\left(M\right)\end{matrix}\right.\)

c) PTHH: \(CaCO_3+2HCl\rightarrow CaCl_2+H_2O+CO_2\)

Theo PTHH: \(n_{HCl}=2n_{CaCO_3}=0,4\left(mol\right)\) \(\Rightarrow m_{ddHCl}=\dfrac{0,4\cdot36,5}{10\%}=146\left(g\right)\)

a.PTHH:CaCl2+Na2CO3--->CaCO3+2NaCl

Ta có:nCaCl2=0,2

=>nCaCO3=nCaCl2=0,2(mol)=>mCaCO3(kết tủa)=100.0,2=2(g)

b.Vdd=100+200=300(ml)=0,3(l)

CM Nacl=(2.0,2)/0,3=4/3(M)(Đề cho 2 chất td vừa đủ nên dd sau pứ chỉ có NaCl)

c.CaCO3+2HCl--->CaCl2+CO2+H2O

nHCl(cần dùng)=2.0.2=0,4(mol)=>mHCl=36,5.0,4=14,6(g)

=>mddHCl=14,6/10%=146(g)

2NaOH + CuSO4 → Cu(OH)2 + Na2SO4

n NaOH = 0,2.5 = 1(mol)

n CuSO4 = 0,1.2 = 0,2(mol)

Ta có :

n NaOH / 2 = 0,5 > n CuSO4 / 1 = 0,2 => NaOH dư

n Cu(OH)2 = n CuSO4 = 0,2 mol

=> m A = 0,2.98 = 19,6 gam

n Na2SO4 = n CuSO4 = 0,2 mol

n NaOH pư = 2n CuSO4 = 0,4(mol)

V dd = 0,2 + 0,1 = 0,3(lít)

Suy ra:

CM Na2SO4 = 0,2/0,3 = 0,67M

CM NaOH = (1 - 0,4)/0,3 = 2M

\(a.n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\\ n_{HCl}=0,2.1,5=0,3\left(mol\right)\\ Mg+2HCl\rightarrow MgCl_2+H_2\\ Vì:\dfrac{0,3}{2}< \dfrac{0,2}{1}\\ \Rightarrow Mgdư\\ n_{H_2}=n_{MgCl_2}=\dfrac{0,3}{2}=0,15\left(mol\right)\\ V_{H_2\left(đktc\right)}=0,15.22,4=3,36\left(l\right)\\ b.V_{ddsau}=V_{ddHCl}=0,2\left(l\right)\\ C_{MddMgCl_2}=\dfrac{0,15}{0,2}=0,75\left(M\right)\)

\(a) n_{Na} = \dfrac{4,6}{23} = 0,2(mol)\\ 2Na + 2H_2O \to 2NaOH + H_2\\ n_{H_2} = \dfrac{1}{2}n_{Na} = 0,1(mol)\\ V_{H_2} = 0,1.22,4 = 2,24(lít)\\ b) n_{NaOH} = n_{Na} = 0,2(mol)\\ V_{dd} = V_{nước} = 0,2(lít)\\ \Rightarrow C_{M_{NaOH}} = \dfrac{0,2}{0,2} = 1M\)

\(n_{NaOH}=0.2\cdot0.5=0.1\left(mol\right)\)

\(n_{CuSO_4}=0.1\cdot2=0.2\left(mol\right)\)

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

\(0.1.............0.05...............0.05...........0.05\)

\(m_{Cu\left(OH\right)_2}=0.05\cdot98=4.9\left(g\right)\)

\(C_{M_{Na_2SO_4}}=\dfrac{0.05}{0.2+0.1}=0.167\left(M\right)\)

\(C_{M_{CuSO_4\left(dư\right)}}=\dfrac{0.2-0.05}{0.1}=1.5\left(M\right)\)

a) \(n_{NaOH}=0,2.1=0,2\left(mol\right);n_{H_2SO_4}=0,15.2=0,3\left(mol\right)\)

PTHH: 2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O

Mol:         0,2                            0,1

Ta có: \(\dfrac{0,2}{2}< \dfrac{0,3}{1}\) ⇒ NaOH hết, H₂SO₄ dư

\(m_{Na_2SO_4}=0,1.142=14,2\left(g\right)\)

b) V_{dd sau pứ} = 0,2 + 0,15 = 0,35 (l)

\(C_{M_{ddNa_2SO_4}}=\dfrac{0,1}{0,35}=\dfrac{2}{7}\approx0,2857M\)

\(C_{M_{ddH_2SO_4dư}}=\dfrac{0,3-0,1}{0,35}=\dfrac{4}{7}\approx0,57M\)