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When silver nitrate is added to an aqueous solution of magnesium chloride, a precipitation reaction occurs that produces silver chloride and magnesium nitrate. When enough AgNO3 is added so that 21.3 g of MgCl2 react, what mass of the AgCl precipitate should form

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sebassandin

Answer:

[tex]m_{AgCl}=64.13gAgCl[/tex]

Explanation:

Hello,

In this case, the undergone chemical reaction turns out:

[tex]2AgNO_3+MgCl_2-->2AgCl+Mg(NO_3)_2[/tex]

In such a way, based on the reacting 21.3 g of magnesium chloride that are consumed the following stoichiometric procedure leads to the required grams of solver chloride precipitate consider their 1 to 2 mole relationship respectively:

[tex]m_{AgCl}=21.3gMgCl_2*\frac{1molMgCl_2}{95.211gMgCl_2}*\frac{2molAgCl}{1molMgCl_2}*\frac{143.32gAgCl}{1molAgCl}=64.13gAgCl[/tex]

Best regards.

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