# Finding the percent by mass of an unknown substance

I'm doing some chemistry homework and a the problem in question goes as follows: "A student performs an experiment to determine the percentage by mass of $$\ce{MgCl2}$$ in a $$\pu{2.94g}$$ sample of a mixture containing anhydrous $$\ce{MgCl2}$$ and $$\ce{KNO3}$$. The student decides to precipitate all of the chloride ion, $$\ce{Cl-}$$, as $$\ce{AgCl}$$, by adding excess queous silver nitrate, $$\ce{AgNO3_{(aq)}}$$ to the mixture sample." $$\pu{5.48g}$$ of $$\ce{AgNO3}$$ are produced, so what is the percent by mass of $$\ce{MgCl2}$$ in the original mixture? I initially just did some basic stoichiometry to find the amount of moles used, but I got $$\pu{0.0439mol}\; \ce{MgCl2}$$ which produces $$\pu{4.17g}$$ which is greater than the $$\pu{2.94g}$$ sample

$$\pu{5.48g}\; \ce{AgCl} \times \pu{1mol}\; \ce{AgCl}\div \pu{64.43g}\; \ce{AgCl} \times \ce{1MgCl2 \div 2AgCl}$$

So I'm simply wondering if I went about solving it wrong or if there's just missing information.