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

can you attach a picture of the lettuce? : )

the answer would be c