Câu 1:
2Fe+3Cl3\(\rightarrow\)2FeCl3
\(n_{Cl_2}=\dfrac{6,72}{22,4}=0,3mol\)
\(n_{FeCl_3}=\dfrac{2}{3}n_{Cl_2}=0,2mol\)
x=0,2.162,5=32,5g
Na2CO3+2HCl\(\rightarrow\)NaCl+CO2+H2O
\(n_{Na_2CO_3}=n_{CO_2}=\dfrac{4,48}{22,4}=0,2mol\)
a=0,2.106=21,2g