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.


There is one error in the question: the precipitate must be AgCl, not AgNO3. I can't quite follow your calculation, but the proper way to do it is:

5.48g AgCl/143 g/mol = 0,04 mol AgCl

1 mol AgCl eqals 0.5 mol MgCl2

0.04/2 = 0.02 mol MgCl2 * 95 g/mol = 1.8g MgCl2

The numbers at rounded off, so look up the exact molecular weights if you want a more precise result.


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