a)
\(CaCO_3+2HCl\rightarrow CaCl_2+CO_2+H_2O\)
\(n_{CaCO_3}=\frac{4}{100}=0,04\left(mol\right)\)
\(n_{CO2}=n_{CaCO3}=0,04\left(mol\right)\)
\(V=0,04.22,4=0,896\left(l\right)\)
\(n_{HCl}=0,04.2=0,08\)
\(m_{dd_{HCl}}=\frac{0,08.36,5}{10\%}=29,2\left(g\right)\)
b)
\(n_{CaCO3}=0,04\left(mol\right)\)
\(n_{HCl}=\frac{7.10\%}{36,5}=0,02\left(mol\right)\)
\(\rightarrow CaCO_3\) dư
\(n_{CO_2}=\frac{n_{HCl}}{2}=0,01\left(mol\right)\)
\(V_{CO2}=0,01.22,4=0,224\left(l\right)\)
Ta có : \(n_{CaCO3}=0,04\left(mol\right)\)
\(a,PTHH:CaCO_3+2HCl\rightarrow CO_2\uparrow+H_2O+CaCl_2\)
_________0,4_______0,8_______0,4_____0,4____0,4_____(mol)
\(V_{CO2}=0,4.22,4=0,896\left(l\right)\)
\(\rightarrow m_{HCl}=0,08.\left(35,5+1\right)=2,92\left(g\right)\)
b, Tự làm nhé
