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.

the grams   of   barium that   are in   2.3 x 10^23   atoms of barium   is   52.4 g (answer c)

    calculation

mass = moles   of ba   x molar   mass of ba

  molar mass of ba from periodic table=   137.3 g/mol

  use the   avogadro's law   constant   to   calculate the moles of barium

  that   is     1 mole =   6.02   x 10^23 atoms

                    ? moles = 2.3 x 10^23   atoms

by cross   multiplication

=(1 mole   x 2.3x 10^23 atoms)   / 6.02   x10^23 atoms)=   0.382   moles

mass   is therefore = 137.3   g/mol x 0.382 mol = 52.4 ( answer c)

explanation:

in chemistry, a solution’s concentration is how much of a dissolvable substance, known as a solute, is mixed with another substance, called the solvent. the standard formula is c = m/v, where c is the concentration, m is the mass of the solute dissolved, and v is the total volume