When ethyl iodide and propyl iodide react with Na in the presence of ether, they form
Three alkanes
Let's recall Wurtz Reaction.
Alkylhalide reacts with metallic sodium in the presence of dry ether to form symmetrical alkanes containing double the number of C atoms present in the alkyl halide.
R–X+2Na+X–RDry−−−→etherR–R+2NaX
CH3–Br+2Na+Br–CH3⟶CH3–CH3+2NaBr
Now,
C2H5l+2Na+C3H7lDry−−−→etherC2H5−C3H7+2Nal
Pentane
C2H5l+2Na+C2H5lDry−−−→etherC2H5–C2H5+2Nal
Butane
C3H7l+2Na+C3H7lDry−−−→EtherC3H7–C3H7+2Nal
Hexane