0.303 g of a compound with the formula mcl2 reacts with aqueous agno3 to yield 0.783 g of a white precipitate of solid agcl according to the equation below. what is the identity of m?

Given:

Mass of the compound MCl2 = 0.303 g

Mass of AgCl formed = 0.783 g

To determine:

The identity of M

Explanation:

Chemical reaction-

MCl2   +   2AgNO3  →   2AgCl  +   M(NO3)2

0.303 g                          0.783 g

Based on the reaction stoichiometry: 1 mole of MCl2 forms  2 moles of AgCl

Molar mass of AgCl = 143 g/mol

# moles of AgCl formed = 0.783/143 = 0.00547 moles

Therefore moles of MCl2 reacted = 0.00547/2 = 0.00274 moles

Molar mass of MCl2 = 0.303/0.00274 = 110.58 g/mol

Atomic mass of M = 110.58 - 2(35.5) = 40.58 g/mol

The element with atomic mass 40 g/mol is calcium Ca

Ans: Identity of M = Ca. The molecule is CaCl2

i think its 100

i hope this !