question about chemistry

Question

When solutions of silver nitrate and calcium chloride are mixed, silver chloride precipitates out of solution according to the equation

2AgNO3(aq)+CaCl2(aq)→2AgCl(s)+Ca(NO3)2(aq)

1.What mass of silver chloride can be produced from 1.15L  of a 0.261M  solution of silver nitrate?

2.The reaction described in Part A required 3.34L of calcium chloride. What is the concentration of this calcium chloride solution?

please tell me how should I solve this

Answer 1

Hi Kal,

First, we need to figure out how many moles of silver nitrate (AgNO₃) we have in 1.15L of a 0.261M. To do that, we must use the equation below:

Number of moles of AgNO₃ = Molarity (moles/L or M) x Volume (L)

                                             n = 0.261 M x 1.15 L

                                                = 0.30015 moles

Now that we have the moles of AgNO₃, we can use it in combination with the balanced equation to determine how much product will be produced. The assumption here is that we have enough of the reactants (in this case it is CaCl₂) and we are only limited by the quantity of AgNO₃.

According to the balanced equation, 2 moles of AgNO₃ produces 2 moles of AgCl, which is the product we care about based on the problem.

Thus, using dimensional analysis:

0.30015 moles of AgNO₃ X 2 moles of AgCl = 0.30015 moles of AgCl

2 moles of AgNO₃

Now we just have to convert these moles into grams, which involves finding the molar mass of AgCl using the periodic table (reminder that molar mass has units of g/mol)

Molar mass of AgCl =143.4 g/mol

So mass of AgCl produced , m = number of moles x molar mass

                                             = 0.30015 mol x 143.4 (g/mol)

                                             = 43.0 g (don't forget about sig figs!)

Hopefully, I got you through part 1 of this problem, and you will be to finish part 2.