[undergrad99]

Help in chemistry. Given the 25 grams of NaCl and 35g AgNO3. Where did they get the 169.9 and 58.4 in the equation. Am I missing something.
25⋅g of sodium chloride are mixed with 35.0⋅g silver nitrate in aqueous solution. How much silver chloride can be prepared?

anor277

Oct 29, 2015

Reaction: Ag+(aq)+Cl−(aq)→AgCl(s)⏐↓

Note that I assume you are using silver nitrate, AgNO3.

Explanation:

Moles of Ag+, 35.0⋅g169.9⋅g⋅mol−1=??

Moles of Cl−, 25.0⋅g58.4⋅g⋅mol−1=??

The Na+(aq) and NO−3(aq) ions are spectators and do not participate in the reaction. Chloride anion is clearly in excess; precipitation AgCl is (almost) quantitative. You are free to do the math.

[undergrad99]

The other half of the problem is.. and where did they get the 143.32??

Explanation:

Balanced Equation

NaCl(aq)+AgNO3(aq)→NaNO3(aq)+AgCl(s)

This is a limiting reagent question. The reactant that produces the least amount of silver chloride is the limiting reagent, and it determines the amount of silver chloride that can be produced.

For each reactant, determine how much silver chloride can be produced.

Determine moles of reactant by dividing given mass of the reactant by its molar mass.

Determine moles silver chloride by multiplying moles of the reactant by the mole ratio between the reactant and silver chloride in the balanced equation, with silver chloride in the numerator.

Determine mass silver chloride by multiplying moles of silver chloride by its molar mass.

Sodium chloride

25.0g NaCl58.44g NaClmol NaCl×1mol AgCl1mol NaCl×143.32g AgCl1mol AgCl=61.3 g AgCl

Silver nitrate

35.0g AgNO3169.87g AgNO3mol AgNO3×1mol AgCl1mol AgNO3×143.32g AgCl1mol AgCl=29.5 g AgCl

AgNO3 is the limiting reagent. The maximum yield of AgCl is 29.5 g.

[Borek]

Have you checked if these are not molar masses?