Bài 1: a) n KOH= 0.5(mol)
n CO2 = 0.5(mol)
2KOH+ CO2times="" new="" roman";mso-hansi-font-family:"times="" roman";="" mso-char-type:symbol;mso-symbol-font-family:wingdings"="">à K2CO3+ H2O (1)
KOH +
CO2à
KHCO3 (2)
Ta có: n KOH/ n CO2= 0.5/ 0.5= 1
=> Xảy ra pt (2)
=> Tạo ra muối KHCO3
n KHCO3= n CO2= 0.5(mol)
m KHCO3= 0.5* 100= 50(g)
b) n CO2= 0.2(mol)
m CO2= 0.2* 44=
8.8(g)
n NaOH= (80* 20%)/
40= 0.4(mol)
2NaOH+ CO2times="" new="" roman";mso-hansi-font-family:"times="" roman";="" mso-char-type:symbol;mso-symbol-font-family:wingdings"="">à Na2CO3+ H2O (1)
NaOH +
CO2à
NaHCO3 (2)
Ta có: n NaOH/ n
CO2= 0.4/ 0.2= 2
=> Xảy ra pt (1)
=> Tạo ra muối Na2CO3
n Na2CO3= n CO2= 0.2(mol)
m Na2CO3= 0.2* 106=
21.2(g)
C% Na2CO3= 21.2/ (80+
8.8) *100%= 23.87(%)
Bài 1: a) n KOH= 0.5(mol)
n CO2 = 0.5(mol)
2KOH+ CO2à K2CO3+ H2O (1)
KOH +
CO2à
KHCO3 (2)
Ta có: n KOH/ n CO2= 0.5/ 0.5= 1
=> Xảy ra pt (2)
=> Tạo ra muối KHCO3
n KHCO3= n CO2= 0.5(mol)
m KHCO3= 0.5* 100= 50(g)
b) n CO2= 0.2(mol)
m CO2= 0.2* 44=
8.8(g)
n NaOH= (80* 20%)/
40= 0.4(mol)
2NaOH+ CO2à Na2CO3+ H2O (1)
NaOH +
CO2à
NaHCO3 (2)
Ta có: n NaOH/ n
CO2= 0.4/ 0.2= 2
=> Xảy ra pt (1)
=> Tạo ra muối Na2CO3
n Na2CO3= n CO2= 0.2(mol)
m Na2CO3= 0.2* 106=
21.2(g)
C% Na2CO3= 21.2/ (80+
8.8) *100%= 23.87(%)
Bài 1: a) n KOH= 0.5(mol)
n CO2 = 0.5(mol)
2KOH+ CO2
times="" new="" roman";mso-hansi-font-family:"times="" roman";="" mso-char-type:symbol;mso-symbol-font-family:wingdings"="">à K2CO3+ H2O (1)
KOH +
CO2à
KHCO3 (2)
Ta có: n KOH/ n CO2= 0.5/ 0.5= 1
=> Xảy ra pt (2)
=> Tạo ra muối KHCO3
n KHCO3= n CO2= 0.5(mol)
m KHCO3= 0.5* 100= 50(g)
b) n CO2= 0.2(mol)
m CO2= 0.2* 44=
8.8(g)
n NaOH= (80* 20%)/
40= 0.4(mol)
2NaOH+ CO2times="" new="" roman";mso-hansi-font-family:"times="" roman";="" mso-char-type:symbol;mso-symbol-font-family:wingdings"="">à Na2CO3+ H2O (1)
NaOH +
CO2à
NaHCO3 (2)
Ta có: n NaOH/ n
CO2= 0.4/ 0.2= 2
=> Xảy ra pt (1)
=> Tạo ra muối Na2CO3
n Na2CO3= n CO2= 0.2(mol)
m Na2CO3= 0.2* 106=
21.2(g)
C% Na2CO3= 21.2/ (80+
8.8) *100%= 23.87(%)