Câu 4 :
\(n_{CaCO3}=\dfrac{60}{100}=0,6\left(mol\right)\)
a) Pt : \(CaCO_3+2HCl\rightarrow CaCl_2+CO_2+H_2O|\)
1 2 1 1 1
0,6 1,2 0,6
b) \(n_{HCl}=\dfrac{0,6.2}{1}=1,2\left(mol\right)\)
⇒ \(m_{HCl}=1,2.36,5=43,8\left(g\right)\)
\(m_{dd}=\dfrac{43,8.100}{20}=219\left(g\right)\)
c) \(n_{CO2}=\dfrac{1,2.1}{2}=0,6\left(mol\right)\)
\(V_{CO2\left(dktc\right)}=0,6.22,4=13,44\left(l\right)\)
Chúc bạn học tốt
a) CaCO3 +2HCl ->CaCl2 +H2O +CO2
b) n CaCO3 =0,6 mol
Theo Pthh tc nHCL=1,2 mol
=>mdd HCl =219 (g)
C) n CO2 = nCaCO3 = 0,6 mol
=> V CO2 =13,44 (L)
