\(a,n_{hhkhí\left(H_2,CO_2\right)}=\dfrac{0,672}{22,4}=0,03\left(mol\right)\\ PTHH:Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\left(1\right)\\ CaCO_3+2HCl\rightarrow CaCl_2+CO_2\uparrow+H_2O\left(2\right)\\b, Theo.pt:n_{HCl}=2n_{hhkhí}=2.0,03=0,06\left(mol\right)\\ V_{ddHCl}=\dfrac{0,06}{0,5}=0,12\left(l\right)=120\left(ml\right)\\ Gọi\left\{{}\begin{matrix}n_{H_2}=a\left(mol\right)\\n_{CO_2}=b\left(mol\right)\end{matrix}\right.\\ ,\dfrac{2a+44b}{a+b}=15.2=30\left(\dfrac{g}{mol}\right)\\ \Rightarrow\left\{{}\begin{matrix}a=0,01\left(mol\right)\\b=0,02\left(mol\right)\end{matrix}\right.\)
\(Theo.pt\left(1\right):n_{CaCO_3}=n_{CaCl_2}=n_{CO_2}=0,02\left(mol\right)\\ Theo.pt\left(2\right):n_{Fe}=n_{FeCl_2}=n_{H_2}=0,01\left(mol\right)\\ m=0,01.56+0,02.100=2,56\left(g\right)\\ c,C_{MCaCl_2}=\dfrac{0,02}{0,12}\approx0,167M\\ C_{MFeCl_2}=\dfrac{0,1}{0,12}\approx0,083M\)
Ôi trời chưa có học cái này !
Theo.pt(1):nCaCO3=nCaCl2=nCO2=0,02(mol)Theo.pt(2):nFe=nFeCl2=nH2=0,01(mol)m=0,01.56+0,02.100=2,56(g)c,CMCaCl2=0,020,12≈0,167MCMFeCl2=0,10,12≈0,083M
