2AgNO3 + MgCl2 ==> 2AgCl + Mg(NO3)2
How many moles AgNO3 do you have? M x L = moles.
How many moles MgCl2 do you have? M x L = moles.
Calculate moles AgCl that will form (using the equation, of course), then convert to grams.
how many grams of a precipitate will form when you mix 45 ml of a 3.5M silver nitrate solution with 45 ml of a 5.4 M magnesium chloride solution
i sorted the problem but have no clue how to start
2 answers
you have 157.5 mmol of AgNO3 and 243 mmol of MgCl2. From the balanced chemical equation, 1 mol of MgCl2 would consume 2 moles of AgNO3. It would then require 315 mmol of AgNO3 to completely consume the MgCl2. Therefore, AgNO3 is the limiting reactant and it would dictate how much AgCl would precipitate out.
Since the ratio of AgCl to AgNO3 is 1:1, multiplying 157.5 mmol with the molecular weight of AgCl (in mg/mmol) would give 26.755g AgCl.
Since the ratio of AgCl to AgNO3 is 1:1, multiplying 157.5 mmol with the molecular weight of AgCl (in mg/mmol) would give 26.755g AgCl.
2 answers