nCaCO3=10/100=0.1(mol)
Ta có hệ phương trình sau
CaCO3 + 2 HCl ➞CaCl2 + CO2 + H2O
0.1................0.2.........0.1.......0.1.......0.1...(mol)
mHCl=0.2*36.5=7.3(g)
C%HCl= \(\dfrac{7.3}{100}*100\)%=7.3%
a=7.3
b)VCO2=0.1*22.4=2.24(l)
c) nNaOH=(50*40/100)/40=0.5(mol)
nOH-=0.5*1=0.5(mol)
Xét\(\dfrac{n OH-}{n CO2}=\dfrac{0.1}{0.5}=0.2<1 \)
=> sau phản ứng OH dư CO2 hết tạo muối Na2CO3
2 NaOH + CO2 ➞ Na2CO3 +H2O
.....0.2.........0.1..........0.1...................(mol)
mNa2CO3=0.1*106=10.6(g)