Question

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

Answer 1

Answer:

$m_{AgCl}=64.13gAgCl$

Explanation:

Hello,

In this case, the undergone chemical reaction turns out:

$2AgNO_3+MgCl_2-->2AgCl+Mg(NO_3)_2$

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:

$m_{AgCl}=21.3gMgCl_2*\frac{1molMgCl_2}{95.211gMgCl_2}*\frac{2molAgCl}{1molMgCl_2}*\frac{143.32gAgCl}{1molAgCl}=64.13gAgCl$

Best regards.