When 80.0 grams of calcium chloride react with silver nitrate, how many grams of silver chloride are produced?
CaCl2 + 2 AgNO3 - Ca(NO3)2 + 2 AgCI

Mass = 206.38 g

Explanation:

Given data:

Mass of calcium chloride = 80.0 g

Mass of silver chloride formed = ?

Solution:

Chemical equation:

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

Number of moles of CaCl₂:

Number of moles = mass/molar mass

Number of moles = 80.0 g/ 110.98 g/mol

Number of moles = 0.72 mol

Now we will compare the moles of CaCl₂ and AgCl

          CaCl₂          :         AgCl

             1               :             2

           0.72          :           2/1×0.72 = 1.44 mol

Mass of silver chloride:

Mass = number of moles × molar mass

Mass = 1.44 mol × 143.32 g/mol

Mass = 206.38 g

consider the reaction below.

 no rx

at 500 k, the reaction is at equilibrium with the following concentrations.

[pci5]= 0.0095 m

[pci3] = 0.020

[ci2] = 0.020 m

what is the equilibrium constant for the given reaction?

0.042

0.42

2.4

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