a, \(MgO+H_2SO_4\rightarrow MgSO_4+H_2O\)

b, Ta có: \(m_{H_2SO_4}=200.9,8\%=19,6\left(g\right)\)

\(\Rightarrow n_{H_2SO_4}=\dfrac{19,6}{98}=0,2\left(mol\right)\)

Theo PT: \(n_{MgO}=n_{MgSO_4}=n_{H_2SO_4}=0,2\left(mol\right)\)

\(\Rightarrow m_{MgO}=0,2.40=8\left(g\right)\)

c, Ta có: m _{dd sau pư} = 8 + 200 = 208 (g)

\(\Rightarrow C\%_{MgSO_4}=\dfrac{0,2.120}{208}.100\%\approx11,54\%\)

\(nAl=\dfrac{2,7}{27}=0,1\left(mol\right)\)

\(mHCl=\dfrac{200.7,3\%}{100\%}=14,6\left(g\right)\)

\(nHCl=\dfrac{14,6}{36,5}=0,4\left(mol\right)\)

PTHH:

\(2Al+6HCl\rightarrow2AlCl_2+3H_2\)

 2         6              2            3    (mol)

0,1       0,3         0,1           0,15   (mol)

LTL : 0,1 / 2 < 0,4/6

=> Al đủ , HCl dư

1. \(VH_2=0,15.22,4=3,36\left(l\right)\)

2. \(mH_2=0,15.2=0,3\left(g\right)\)

mdd = mAl + mddHCl - mH2 = 2,7 + 200 - 0,3 = 202,4 (g)

\(mH_2SO_{4\left(dưsaupứ\right)}=0,1.98=9,8\left(g\right)\)

\(mAlCl_2=0,1.98=9,8\left(g\right)\)

\(C\%_{ddH_2SO_4}=\dfrac{9,8.100}{202,4}=4,84\%\)

\(C\%_{AlCl_2}=\dfrac{9,8.100}{202,4}=4,84\%\)

nNaOH = 4/40 = 0.1 (mol)

PTHH: NaOH + HCl -> NaCl + H2O

Từ PTHH: nNaCl = nHCl = nNaOH = 0.1 (mol)

a) mNaCl = 0.1*(23+35.5) = 5.85(g)

b) mHCl = 0.1*(1+35.5) = 3.65(g)

C%ddHCl = 3.65/100 * 100% = 3.65%

\(n_{BaCl_2}=\dfrac{31,2}{208}=0,15mol\)

\(BaCl_2+H_2SO_4\rightarrow BaSO_4\downarrow+2HCl\)

  0,15        0,15           0,15           0,3

a)\(m_{BaSO_4}=0,15\cdot233=34,95\left(g\right)\)

b)\(m_{H_2SO_4}=0,15\cdot98=14,7\left(g\right)\)

    \(\Rightarrow m_{ddH_2SO_4}=\dfrac{14,7}{19,6}\cdot100=75\left(g\right)\)

c)\(m_{HCl}=0,3\cdot36,5=10,95\left(g\right)\)

  \(m_{ddsau}=31,2+75-34,95=71,25\left(g\right)\)

  \(\Rightarrow C\%_{HCl}=\dfrac{10,95}{71,25}\cdot100\%=15,37\%\)

Cho Ba(OH)₂ vào muối Al sẽ có 2TH sau:

TH1: kết tủa Al(OH)₃ chưa bị hòa tan

Al³⁺ + 3OH^– → Al(OH)₃↓

→ n_Al(OH)3 = n_Al3+ → n_Al(OH)3 = xn + 0,04n

TH2: kết tủa Al(OH)₃ bị hòa tan một phần

Al³⁺                 + 3OH^–                → Al(OH)₃↓

(xn + 0,04n)→ 3(xn + 0,04n)    (xn + 0,04n)

Al(OH)₃                 + OH^–           → AlO₂^– + 2H₂O

0,952 – 3(xn + 0,04n) ←0,952

→ n_Al(OH)3 = 4xn + 0,16n – 0,952

mNaOH=11(g) -> nNaOH= 0,275(mol)

mH3PO4=9,8(g) -> nH3PO4=0,1(mol)

Ta có: 2< nNaOH/nH3PO4 = 0,275/0,1=2,75< 3 

=> P.ứ kết thúc thu được hỗn hợp dd Na3PO4 và Na2HPO4

PTHH: 3 NaOH + H3PO4 -> Na3PO4 + 3 H2O

3x_____________x________x(mol)

2 NaOH + H3PO4 -> Na2HPO4 +2 H2O

2y_____y__________y(mol)

mddX=55+24,5=79,5(g)

\(\left\{{}\begin{matrix}3x+2y=0,275\\x+y=0,1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,075\\y=0,025\end{matrix}\right.\)

=> mNa3PO4=0,075.164=12,3(g)

mNa2HPO4=142.0,025=3,55(g)

=>C%ddNa3PO4=(12,3/79,5).100=15,472%

C%ddNa2HPO4=(3,55/79,5).100=4,465%

Ta có: \(\left\{{}\begin{matrix}n_{NaOH}=\dfrac{55\cdot20\%}{40}=0,275\left(mol\right)\\n_{H_3PO_4}=\dfrac{24,5\cdot40\%}{98}=0,1\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\) Tạo HPO₄^2- và PO₄^3-

PTHH: \(2NaOH+H_3PO_4\rightarrow Na_2HPO_4+2H_2O\)

                     2a_______a___________a_______2a   (mol)

           \(3NaOH+H_3PO_4\rightarrow Na_3PO_4+3H_2O\)

                     3b_______b_________b______3b    (mol)

Ta lập HPT: \(\left\{{}\begin{matrix}2a+3b=0,275\\a+b=0,1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=0,025\\b=0,075\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}C\%_{Na_2HPO_4}=\dfrac{0,025\cdot142}{55+24,5}\cdot100\%\approx4,47\%\\C\%_{Na_3PO_4}=\dfrac{0,075\cdot164}{55+24,5}\cdot100\%\approx15,47\%\end{matrix}\right.\)

Ta có: \(n_{CuO}=\dfrac{40}{80}=0,5\left(mol\right)\)

PT: \(CuO+2HCl\rightarrow CuCl_2+H_2O\)

____0,5____________0,5 (mol)

a, \(m_{CuCl_2}=0,5.135=67,5\left(g\right)\)

b, Có: m _{dd sau pư} = m_CuO + m _{dd HCl} = 40 + 200 = 240 (g)

\(\Rightarrow C\%_{CuCl_2}=\dfrac{67,5}{240}.100\%=28,125\%\)

Bạn tham khảo nhé!