What mass of silver chloride can be produced from 1.30 L of a 0.245 M solution of silver nitrate? The reaction described in Part A required 3.36 L of calcium chloride. What is the concentration of this calcium chloride solution?

Please explain, I don't know where to begin.

Explanation:

Given parameters:

Volume of AgNO₃ = 1.3L

Molarity or concentration AgNO₃   = 0.245M

Unknown:

Mass of AgCl produced = ?

Solution:

To solve this problem, we have to work from the known compound to the unknown one using the mole concept.

The known compound is the one that we can obtain the value of the number of moles from. Here it is the given AgNO₃ that can furnish us with this piece of information;

     Now let us establish the balanced reaction equation;

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

Now;

Find the number of moles of AgNO₃;

 Number of moles of AgNO₃  = concentration x volume  

                                                  = 1.3 x 0.245

                                                   = 0.32moles

From the balanced reaction equation;

        2 mole of AgNO₃  produced 2 moles of AgCl

        0.32 moles of AgNO₃ will produce 0.32moles of AgCl

Now that we know the number of moles of the AgCl, we can find the mass;

         Mass of AgCl  = number of moles x molar mass

Molar mass of AgCl  = 107.9 + 35.5  = 143.4g/mol

  Mass of AgCl  = 0.32 x 143.4  = 45.89g

catya that's not even a b or c

ca has a crystalline lattice with positive ions (since it is a metal, metals are positively charged) and electrons flow freely in its structure since the ca atoms have a large nuclear radius so the electrons are not tightly held, therefore the electrons can flow freely since they are not being highly attracted by the nucleus. moreover, one of the properties of metals are that they are conductors of electricity, thus proving ca is the element.