\(n_{AgNO_3}=0,25\cdot0,2=0,05\left(mol\right);n_{MgCl_2}=0,1\cdot0,3=0,03\left(mol\right)\\ a,PTHH:2AgNO_3+MgCl_2\rightarrow2AgCl\downarrow+Mg\left(NO_3\right)_2\\ \text{Vì }\dfrac{n_{AgNO_3}}{2}< \dfrac{n_{MgCl_2}}{1}\text{ nên sau phản ứng }MgCl_2\text{ dư}\\ \Rightarrow n_{AgCl}=n_{AgNO_3}=0,05\left(mol\right)\\ \Rightarrow a=m_{AgCl}=0,05\cdot143,5=7,175\left(g\right)\\ 2,n_{Mg\left(NO_3\right)_2}=\dfrac{1}{2}n_{AgNO_3}=0,025\left(mol\right)\\ \Rightarrow C_{M_{Mg\left(NO_3\right)_2}}=\dfrac{0,025}{0,2+0,3}=0,05M\)

Đáp án:

  CÂU 3:

 PTHH: 

nAgNO3=0,2×0,25=0,05(mol)nAgNO3=0,2×0,25=0,05(mol)

nMgCl2=0,3×0,1=0,03(mol)nMgCl2=0,3×0,1=0,03(mol)

 Xét nMgCl21nMgCl21

→  hết,  dư.

Tính theo số mol 

→ nMg(NO3)2=12×0,05=0,025(mol)nMg(NO3)2=12×0,05=0,025(mol)

⇒ a=mAgCl=0,05×143,5=7,175(g)a=mAgCl=0,05×143,5=7,175(g)

 - Dung dịch  gồm:  dư và 

Xem như thể tích dung dịch sau phản ứng thay đổi không đáng kể.

→ Vdd=0,2+0,3=0,5(l)Vdd=0,2+0,3=0,5(l)

⇒ C(M)Mg(NO3)2=0,0250,5=0,05(M)

a, \(n_{HCl}=0,2.0,1=0,02\left(mol\right)=n_{H^+}=n_{Cl^-}\)

\(n_{H_2SO_4}=0,2.0,15=0,03\left(mol\right)=n_{SO_4^{2-}}\) \(\Rightarrow n_{H^+}=2n_{H_2SO_4}=0,06\left(mol\right)\)

\(\Rightarrow\Sigma n_{H^+}=0,02+0,06=0,08\left(mol\right)\)

\(n_{Ba\left(OH\right)_2}=0,3.0,05=0,015\left(mol\right)=n_{Ba^{2+}}\) 

\(\Rightarrow n_{OH^-}=2n_{Ba\left(OH\right)_2}=0,03\left(mol\right)\)

\(H^++OH^-\rightarrow H_2O\)

0,03___0,03 (mol) ⇒ n_H⁺ dư = 0,05 (mol)

\(Ba^{2+}+SO_4^{2-}\rightarrow BaSO_4\)

0,015___0,015______0,015 (mol) ⇒ n_SO4^2- dư = 0,015 (mol)

⇒ m = m_BaSO4 = 0,015.233 = 3,495 (g)

\(\left[Cl^-\right]=\dfrac{0,02}{0,2+0,3}=0,04\left(M\right)\)

\(\left[H^+\right]=\dfrac{0,05}{0,2+0,3}=0,1\left(M\right)\)

\(\left[SO_4^{2-}\right]=\dfrac{0,015}{0,2+0,3}=0,03\left(M\right)\)

b, pH = -log[H⁺] = 1

\(n_{FeCl_3}=0.1\cdot0.1=0.01\left(mol\right)\)

\(n_{NaOH}=0.5\cdot0.1=0.05\left(mol\right)\)

\(FeCl_3+3NaOH\rightarrow Fe\left(OH\right)_3+3NaCl\)

\(1............3\)

\(0.01...........0.05\)

Lập tỉ lệ : \(\dfrac{0.01}{1}< \dfrac{0.05}{3}\Rightarrow NaOHdư\)

Các chất có trong D : \(NaCl:0.03\left(mol\right),NaOH\left(dư\right):0.02\left(mol\right)\)

\(V=0.1+0.5=0.6\left(l\right)\)

\(\left[Na^+\right]=\dfrac{0.03+0.02}{0.06}=\dfrac{1}{12}\left(M\right)\)

\(\left[Cl^-\right]=\dfrac{0.03}{0.06}=0.5\left(M\right)\)

\(\left[OH^-\right]=\dfrac{0.02}{0.6}=\dfrac{1}{30}\left(M\right)\)

\(b.\)

\(m_{Fe\left(OH\right)_3}=0.01\cdot107=1.07\left(g\right)\)