a,\(m_{H_2SO_4}=49\%.200=98\left(g\right)\Rightarrow n_{H_2SO_4}=\dfrac{98}{98}=1\left(mol\right)\)

PTHH: Zn + H₂SO₄ → ZnSO₄ + H₂

Mol:     1            1                1             1

⇒ m = 1.65 = 65 (g)

b, \(V_{H_2}=1.24=24\left(l\right)\)

c, \(m_{ZnSO_4}=1.161=161\left(g\right)\)

   m_{dd sau pứ} = 65+200-1.2=263 (g)

\(\Rightarrow C\%_{ddZnSO_4}=\dfrac{161.100\%}{263}=61,22\%\)

PTHH:

Ba+H₂SO₄➞BaSO₄↓+H₂↑(1)

0,1 0,1 0,1 0,1

Ba+2H₂O➞ Ba(OH)₂+ H₂(2)

0,1 0,1 0,1

Ta có:{ n_Ba=\(\frac{27,4}{137}=0,2\left(mol\right)\);\(n_{H_2SO_4}=\frac{100.9,8\%}{98}=0,1\left(mol\right)\)

Lập tỉ lệ : \(\frac{0,2}{1}>\frac{0,1}{1}\)

➞ Ở phương trình 1, Ba dư ,H₂SO₄ hết

➞ Xảy ra phản ứng(2)

Đặt các số mol trên phương trình ta có:

➞\(V_{H_2}\left(đktc\right)=\left(0,1+0,1\right).22,4=4,48\left(l\right)\)

Lại có \(m_{dd}\) sau phản ứng=\(27,4+100-0,1.233-0,2.2=103,7\left(g\right)\)

➞\(C\%_{Ba\left(OH\right)_2}=\frac{0,1.171}{103,7}.100\%=16,4899\%\)

Ta có: \(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)

a. PTHH: Zn + H₂SO₄ ---> ZnSO₄ + H₂

b. Theo PT: \(n_{H_2}=n_{Zn}=0,1\left(mol\right)\)

=> \(V_{H_2}=0,1.22,4=2,24\left(lít\right)\)

c. Theo PT: \(n_{H_2SO_4}=n_{Zn}=0,1\left(mol\right)\)

=> \(m_{H_2SO_4}=0,1.98=9,8\left(g\right)\)

Ta có: \(C_{M_{H_2SO_4}}=\dfrac{9,8}{m_{dd_{H_2SO_4}}}.100\%=25\%\)

=> \(m_{dd_{H_2SO_4}}=39,2\left(g\right)\)

Ta có: \(m_{H_2}=0,1.2=0,2\left(g\right)\)

=> \(m_{dd_{ZnSO_4}}=6,5+39,2-0,2=45,5\left(g\right)\)

Theo PT: \(n_{ZnSO_4}=n_{Zn}=0,1\left(mol\right)\)

=> \(m_{ZnSO_4}=0,1.161=16,1\left(g\right)\)

=> \(C_{\%_{ZnSO_4}}=\dfrac{16,1}{45,5}.100\%=35,4\%\)

Câu 1:

a, \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

b, \(n_{Zn}=\dfrac{16,25}{65}=0,25\left(mol\right)\)

\(n_{H_2}=n_{Zn}=0,25\left(mol\right)\)

\(\Rightarrow V_{H_2}=0,25.24,79=6,1975\left(l\right)\)

c, \(n_{HCl}=2n_{Zn}=0,5\left(mol\right)\)

\(\Rightarrow m_{HCl}=0,5.36,5=18,25\left(g\right)\)

\(\Rightarrow m_{ddHCl}=\dfrac{18,25}{10\%}=182,5\left(g\right)\)

d, \(n_{ZnCl_2}=n_{Zn}=0,25\left(mol\right)\)

\(\Rightarrow C\%_{ZnCl_2}=\dfrac{0,25.136}{16,25+182,5-0,25.2}.100\%\approx17,15\%\)

Câu 2:

a, \(2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\)

c, \(n_{NaOH}=\dfrac{40}{40}=1\left(mol\right)\)

Theo PT: \(n_{H_2SO_4}=\dfrac{1}{2}n_{NaOH}=0,5\left(mol\right)\)

\(\Rightarrow V_{H_2SO_4}=\dfrac{0,5}{2}=0,25\left(l\right)\)

d, \(n_{Na_2SO_4}=\dfrac{1}{2}n_{NaOH}=0,5\left(mol\right)\)

\(\Rightarrow C_{M_{Na_2SO_4}}=\dfrac{0,5}{0,25}=2\left(M\right)\)

a) \(n_{Na_2CO_3}=\dfrac{21,2}{106}=0,2\left(mol\right)\)

PTHH: Na₂CO₃ + 2HCl --> 2NaCl + CO₂ + H₂O

                0,2------------------>0,4---->0,2

m_{dd sau pư} = 200 + 21,2 - 0,2.44 = 212,4(g)

=> \(C\%\left(NaCl\right)=\dfrac{0,4.58,5}{212,4}.100\%=11,017\%\)

$PTHH:Na_2CO_3+2HCl\to 2NaCl+H_2O+CO_2\uparrow$

$n_{Na_2CO_3}=\dfrac{21,2}{106}=0,2(mol)$

Theo PT: $n_{NaCl}=n_{CO_2}=0,2(mol)$

$\Rightarrow m_{NaCl}=0,4.58,5=23,4(g);m_{CO_2}=0,2.44=8,8(g)$

$\Rightarrow C\%_{NaCl}=\dfrac{23,4}{21,2+200-8,8}.100\%\approx 11,01\%$

$n_{Al_2O_3} = \dfrac{4,08}{102} = 0,04(mol)$

\(Al_2O_3+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2O\)

0,04          0,12                    0,04                                (mol)

$m_{Al_2(SO_4)_3} = 0,04.342 = 13,68(gam)$

$C\%_{H_2SO_4} = \dfrac{0,12.98}{280}.100\% = 4,2\%$

a) \(n_{HCl}=\dfrac{m}{M}=\dfrac{13}{65}=0,2\left(mol\right)\)

\(Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\) 
   1       2           1        1    (mol)

0,2      0,4        0,2     0,2   (mol)

b) Thể tích khí hidro:

V = n.22,4 = 0,2.22,4 = 4,48 (l)

c) Khối lượng muối tạo thành:

\(m_{ZnCl_2}=n.M=0,2.\left(65+35,5.2\right)=27,2\left(g\right)\)

d) \(m_{ctHCl}=n.M=0,4.\left(1+35,5\right)=14,6\left(g\right)\)

\(C\%_{HCl}=\dfrac{m_{ctHCl}}{m_{ddHCl}}.100\%=\dfrac{14,6}{200}.100\%=7,3\%\)

a) Zn + 2HCl → ZnCl₂ + H₂↑

b)