\(n_{BaSO_4}=\frac{m}{M}=\frac{32,62}{233}=0,14mol\)

PTHH:

\(Ba\left(OH\right)_2+H_2SO_4\rightarrow BaSO_4+2H_2O\)

0,14 0,14 0,14 0,28 (mol)

Gọi \(V_{ddH_2SO_4}\)cần thêm là x

\(n_{H_2SO_4}=\frac{m}{M}=\frac{98}{98}=1mol\)

\(C^{\left(A\right)}_{M_{H_2SO_4}}=\frac{1}{1}=1M\)

\(n^{\left(A\right)}_{H2SO4}=C_M.V=1.x=xmol\)

\(n_{H2SO4}=C_M.V=2.0,4=0,8mol\)

\(C_{MX}=\frac{n}{V}=\frac{0,8+x}{0,4+x}\left(M\right)\)

\(n_X=C_{MX}.V\)

\(\Leftrightarrow0,14=\frac{0,8+x}{0,4+x}.0,1\)

\(\Leftrightarrow\frac{0,14}{0,1}=\frac{0,8+x}{0,4+x}\)

⇔0,08+0,1x=0,56+0,14x

⇔x=0,6(l)

Vậy cần thêm 0,6 l dung dịch

\(a.n_{HCl}=0,1.1=0,1\left(mol\right)\\ n_{H_2SO_4}=0,5.0,1=0,05\left(mol\right)\\ \left[HCl\right]=\dfrac{0,1}{0,1+0,1}=0,5\left(M\right)\\ \left[H_2SO_4\right]=\dfrac{0,05}{0,1+0,1}=0,25\left(M\right)\\ \left[H^+\right]=0,5+0,25.2=1\left(M\right)\\ \left[SO^{2-}_4\right]=\left[H_2SO_4\right]=0,25\left(M\right)\\ \left[Cl^-\right]=\left[HCl\right]=0,5\left(M\right)\)

\(b.BaCl_2+H_2SO_4\rightarrow BaSO_4\downarrow+2HCl\\ n_{BaSO_4}=n_{H_2SO_4}=0,05\left(mol\right)\\ m_{\downarrow}=m_{BaSO_4}=233.0,05=11,65\left(g\right)\)

\(n_{H^+}=n_{HNO_3}=V\)mol

\(n_{OH^-}=n_{NaOH}=0,5.0,2=0,1\) mol

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

0,1<--0,1

\(\Rightarrow n_{H^+}=V=0,1\)lít = 100 ml

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

0,1 -----> 0,1 ---------->0,1

\(NaNO_3\rightarrow Na^++NO_3^-\)

\(\Rightarrow\left[Na^+\right]=\left[NO_3^-\right]=\dfrac{0,1}{0,1+0,2}=0,33M\)

co 2A+2x HCL ->2ACL_x+xH₂(1)

mdd_HCL=100*1.05=105(g)

n_HCL=0.1*0.1=0.01(mol)

tu(1)->n_h2=1/2*0.01=0.005(mol);

mh2=0.01(mol)

co mddsau=mdddau+m-mh2

->m=0.12(g);

nA=0.12/MA

tu(1)->nA=1/xnHCL=1/x*0.01

=>(tu lam iep)

Gọi CTTQ: M₂O₃

m_H2SO4 = \(\dfrac{9,8\times300}{100}=29,4\left(g\right)\)

n_H2SO4 = \(\dfrac{29,4}{98}=0,3\left(mol\right)\)

Pt: M₂O₃ + 3H₂SO₄ --> M₂(SO₄)₃ + 3H₂O

..0,1 mol<--0,3 mol----> 0,1 mol

Ta có: 10,2 = 0,1.(2M + 48)

=> M = 27 (Al)

Vậy M là Nhôm (Al). CTHH: Al₂O₃

mdd sau pứ = moxi + mdd H2SO4 = 10,2 + 300 = 310,2 (g)

C% dd Al₂(SO₄)₃ = \(\dfrac{0,1\times342}{310,2}.100\%=11,025\%\)

3. a) AgNO3 +HCl --> AgCl +HNO3 (1)

nHCl=0,4(mol)=>mHCl=14,6(g)

nAgNO3=0,3(mol)

lập tỉ lệ :

\(\dfrac{0,4}{1}>\dfrac{0,3}{1}\)

=>HCl dư ,AgNO3 hết => bài toán tính theo AgNO3

theo (1) : nHCl(dư)=nHNO3=nAgCl=nAgNO3=0,3(mol)

=>mAgCl=43,05(g)

b)mdd sau pư=14,6+300-43,05=271,55(g)

mHCl(dư)=3,65(g)

mHNO3=18,9(g)

=>C%dd HNO3=6,96(%)

C%dd HCl dư=1,344(%)

2. a) Mg +2HCl --> MgCl2 +H2 (1)

nH2=0,3(mol)

theo (1) : nMg=nH2=0,3(mol)

=>mMg=7,2(g)=>mCu=4,8(g)

=>nCu=0,075(mol)

%mMg=60(%)

%mCu=40(%)

b) theo (1) : nHCl=2nH2=0,6(mol)

=>mdd HCl=100(g)

c) mH2=0,6(mol)

mdd sau pư= 7,2+100-0,6=106,6(g)

theo (1) : nMgCl2=nMg=0,3(mol)

=>mMgCl2=28,5(g)

=>C%dd MgCl2=26,735(%)

a)Na2CO3 +2HCl --> 2NaCl + CO2 +H2O (1)

nCO2=0,02(mol)

theo (1) : nNaCl=nHCl =2nCO2=0,04(mol)

=>C_{Mdd HCl}=0,04/0,02=2(M)

b) theo (1) : nNa2CO3=nCO2=0,02(mol)

=>mNa2CO3=2,12(g)

=>mNaCl (hh bđ) =2,88(g)

mNaCl(sau pư)=0,04.58,5=2,34(g)

=>m_{muối tạo thành sau pư}=5,22(g)

c) %mNa2CO3=42,4(%)

%mNaCl(hh bđ)=57,6(%)

Gọi n_Fe=n_R= x (mol)

Ta thấy cả Fe và R khi tác dụng với HCl đều đưa về muối clorua hóa trị II

Tổng quát

\(A+2HCl\rightarrow ACl_2+H_2\)

Ta có : m_Muối= m_A +m_Cl^- => m_Cl^-=7,1 (g) => n_Cl^-=0,2mol

mà n_Cl^-=n_HCl=2n_H2=2n_A=>\(\left\{{}\begin{matrix}n_{HCl}=0,2\left(mol\right)\\n_{H2}=0,1\left(mol\right)\\n_A=0,1\left(mol\right)=2x\end{matrix}\right.\Rightarrow x=0,05\left(mol\right)\)

a. V= 22,4.0,1=2,24(l)

\(\left[HCl\right]=\dfrac{0,2}{0,1}=2M\)

b. Ta có : m_A= m_Fe + m_R= 0,05.56 + 0,05.M_R= 4 => M_R=24(g/mol)

=> R là Mg