Considering the reaction stoichiometry, you obtain:
The balanced reaction is:
PbĀ²āŗ(aq) + 2 KCl(aq) ā PbClā(s) + 2 Kāŗ(aq)
By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:
The molar mass of each compound or ion is:
Then, by reaction stoichiometry, the following amount of mass of each compound participate in the reaction:
A sample of water was found to contain 2.6x10ā»Ā³ mole PbĀ²āŗ.
a) Then you can apply the following rule of three: Ā if by stoichiometry 2 moles of PbĀ²āŗ react with 149.1 grams of KCl, 2.6x10ā»Ā³ moles of PbĀ²āŗ react with how much mass of KCl?
[tex]mass of KCl=\frac{2.6x10^{.3}moles of Pb^{2+} x149.1 grams of KCl }{2moles of Pb^{2+} }[/tex]
mass of KCl= 0.19383 grams
Finally, 2.6x10ā»Ā³ moles of PbĀ²āŗ react with 0.19383 grams of KCl.
b) Then you can apply the following rule of three: Ā If by stoichiometry 2 moles of PbĀ²āŗ produces 278.1 grams of PbClā, 2.6x10ā»Ā³ moles of PbĀ²āŗ produces how much mass of PbClā?
[tex]mass of PbCl_{2} =\frac{2.6x10^{.3}moles of Pb^{2+} x278.1 grams of PbCl_{2} }{2moles of Pb^{2+} }[/tex]
mass of PbClā= 0.36153 grams
Finally, 2.6x10ā»Ā³ moles of PbĀ²āŗ produces 0.36153 grams of PbClā.
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