Câu trả lời:
a, nCO2=\(\dfrac{V}{22,4}\)=\(\dfrac{1,344}{22,4}\)=0,06(mol)
PTHH: CO2+2KOH --->K2CO3+H20
0,06 ---------------->0,06
b, nK2CO3=0.06 mol=>mK2CO3=n.M=0,06.138=8,28(g)
c,PTHH:K2CO3+2HCl --->2KCl+CO2+H2O
0,12 <----------- 0.06
nHCl=0,12 mol =>mHCl=n.M=0,12.36,5=4,38(g)