Question

Answer 1

Gọi số mol của mỗi chất trong hỗn hợp X là x mol
Ta có : 84x + 100x + 111x + 208x = 37,725
=> x=0,075 (mol)
n_Na2O =\(\dfrac{4,65}{62}\) =0,075 (mol)

Na2O + H2O\(\rightarrow\) 2NaOH
0,075 .................0,15 (mol)
NaHCO3 + NaOH \(\rightarrow\) Na2CO3 + H20
0,075...........0,075........0,075 (mol)
2KHCO3 + 2NaOH \(\rightarrow\) K2CO3 + Na2CO3 + 2H20
0,075...........0,075........0,0375........0,0375
CaCl2 + Na2CO3 \(\rightarrow\) CaCO3 + 2NaCl
0,075 ......0,075 ..........0,075.........0,15 (mol)

Na2CO3+ BaCl2 →BaCO3 +2NaCl

0,0375.......0,0375....0,0375....0,075 (mol)
BaCl2 + K2CO3 \(\rightarrow\) BaCO3 + 2KCl
0,0375.....0,0375.........0,0375.......0,075 (mol)
m_ddsaupu= 37,725 + 130 +4,65 -(100.0,075 +197.0,0375.2) =150,1g
\(C\%_{NaCl}=\dfrac{\left(0,15+0,075\right).58,5}{150,1}.100=8,77\%\)

\(C\%_{KCl}=\dfrac{0,075.74,5}{150,1}.100=3,72\%\)