Question

Answer 1

Gọi \(\left\{{}\begin{matrix}n_{K_2CO_3}=a\left(mol\right)\\n_{KCl}=b\left(mol\right)\\n_{KHCO_3}=c\left(mol\right)\end{matrix}\right.\)

=> 138a + 74,5b + 100c = 39,09 (1)

PTHH: \(K_2CO_3+2HCl\rightarrow2KCl+CO_2+H_2O\)

                  a------->2a------->2a----->a

             \(KHCO_3+HCl\rightarrow KCl+CO_2+H_2O\)

                   c-------->c-------->c------>c

=> \(a+c=\dfrac{6,72}{22,4}=0,3\) (2)

dd sau pư chứa \(\left\{{}\begin{matrix}KCl:2a+b+c\left(mol\right)\\HCl_{dư}:x\left(mol\right)\end{matrix}\right.\)

=> Mỗi phần chứa \(\left\{{}\begin{matrix}KCl:a+0,5b+0,5c\left(mol\right)\\HCl_{dư}:0,5x\left(mol\right)\end{matrix}\right.\)

- Phần 1:

n_NaOH = 0,25.0,4 = 0,1 (mol)

PTHH: \(NaOH+HCl\rightarrow NaCl+H_2O\)

                0,1---->0,1

=> 0,5x = 0,1

=> x = 0,2 

- Phần 2:

\(n_{AgCl}=\dfrac{51,66}{143,5}=0,36\left(mol\right)\)

PTHH: \(KCl+AgNO_3\rightarrow AgCl\downarrow+KNO_3\)

  (a+0,5b+0,5c)-------->(a+0,5b+0,5c)

            \(HCl+AgNO_3\rightarrow AgCl\downarrow+HNO_3\)

               0,1--------------->0,1

=> a+0,5b+0,5c + 0,1 = 0,36

=> a + 0,5b + 0,5c = 0,26 (3)

(1)(2)(3) => a = 0,2 (mol); b = 0,02 (mol); c = 0,1 (mol)

\(\left\{{}\begin{matrix}m_{K_2CO_3}=0,2.138=27,6\left(g\right)\\m_{KCl}=0,02.74,5=1,49\left(g\right)\\m_{KHCO_3}=0,1.100=10\left(mol\right)\end{matrix}\right.\)

n_HCl(bđ) = 2a + c + x = 0,7 (mol)

=> m_HCl(bd) = 0,7.36,5 = 25,55 (g)

=> \(m_{dd.HCl}=\dfrac{25,55.100}{10,52}\approx242,87\left(g\right)\)

=> \(V_{dd.HCl}=\dfrac{242,87}{1,05}\approx231,3\left(ml\right)\)