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 34.3 g of MgCl2 react, what mass of the AgCl precipitate should form

103.62 g of AgCl.

Explanation:

Step 1:

The balanced equation for the reaction. This is given below:

2AgNO3 + MgCl2 —> 2AgCl + Mg(NO3)2

Step 2:

Determination of the mass of MgCl2 that reacted and the mass of AgCl produced from the balanced equation.

This is illustrated below:

Molar mass of MgCl2 = 24 + (2x35.5) = 95 g/mol

Mass of MgCl2 from the balanced equation = 1 x 95 = 95 g

Molar mass of AgCl = 108 + 35.5 = 143.5 g/mol

Mass of AgCl from the balanced equation = 2 x 143.5 = 287 g

Thus, from the balanced equation above,

95 g of MgCl2 reacted to produce 287 g of AgCl.

Step 3:

Determination of the mass of AgCl produced from the reaction of 34.3 g of MgCl2.

The mass of AgCl produced from the reaction can be obtained as follow:

Form the balanced equation above,

95 g of MgCl2 reacted to produce 287 g of AgCl.

Therefore, 34.3 g of MgCl2 will react to produce = (34.3 x 287)/95 = 103.62 g of AgCl.

Therefore, 103.62 g of AgCl were produced from the reaction.

i believe the answer is 5,000