Question

Part A--a student attempts to determine the formula of an unknown chloride salt , MCl A student first takes a 9.79 g sample of the salt and fully decomposes it. The mass of chlorine gas produced is 7.29 g What is the mass of metal M contained in the salt? The mass of metal M collected from part A is then reactedwith Zinc nitrate and 6.77 zinc metal is produced. how many moes of metal M reacted??M+Zn(NO3)2 M(NO3)2+Zn What is the molar mas and the identity of metal M? What is the emporical formula of the original chloride slat in part A What is the

Answer 1

mass of the salt(MClx) = 9.79 g

mass of cl2 liberated by decomposition = 7.29 g

mass of metal present in salt(MClx) = 9.79-7.29 = 2.5 g

M+Zn(NO3)2 ---> M(NO3)2+Zn

No of mol of Zn produced = w/Mwt = 6.77/65.4 = 0.1035 mol

from the above reaction,

No of mol of Zn(NO3)2 reacted = 0.1035 mol

No of mol of Metal(M) reacted = 0.1035 mol

molar mass of the metal(M) = w/n = 2.5/0.1035 = 24.15 g/mol

the identity of metal with molarmass of 24 g/mol = Magnesium(Mg)

so that, The emporical formula of the original chloride salt = MgCl2