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

How many grams of AgCi are formed when 0.200 L of 0.200 M AGNO3 reacts with an excess of CaCl2?

5.732 grams of AgCl is formed when 0.200 L of 0.200 M AGNO3 reacts with an excess of CaCl2.

Explanation:

The balanced equation:

2 AgNO3(aq) + CaCl2(aq) -----> 2 AgCl(s) + Ca(NO3)2(aq)

data given:

volume of AgNO3 = 0.2 L

molarity of AgNO3 = 0.200 M

atomic weight of AgCl= 143.32 gram/mole

from the formula, number of moles can be calculated

Molarity =

number of moles of AgNO3 = 0.04

From the reaction:

2 moles of AgNO3 reacts to form 2 moles of AgCl

0.04 moles of AgNO3 reacts to form x mole of AgCl

=

= 0.04 moles of AgCl is formed

mass of AgCl formed is calculated by multiplying number of moles with atomic mass of AgCl

mass of AgCl = 0.04 x 143.32

                       = 5.732 grams of AgCl is formed.

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false

because it does not have the potential to be entirely used up