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Sagot :
2.3352 g of PbCl₂ are formed when 50. 0 ml of 0. 336 m KCl react with pb(no3)2.
The balanced equation for the above double displacement reaction is as follows;
2KCl + Pb(NO₃)₂ ---> PbCl₂ + 2KNO₃
Stoichiometry of KCl to PbCl₂ is 2:1
This means that 2 mol of KCl would react with every 1 mol of PbCl₂
The molarity of KCl = 0. 336 M
in 1 L of KCl, there are mol
Therefore in 50. 0 ml of KCl, there are= [tex]\frac{0. 336 * 50}{1000}[/tex]
Number of KCl moles reacted = 0.0168 mol
according to stoichiometry
number of PbCl₂ moles formed = 1/2 x number of KCl moles reacted
Therefore number of PbCl₂ moles formed = 0.0168 mol/2 = 0.0084 mol
molar mass of PbCl₂ = 278 g/mol
mass of PbCl₂ formed = 278 g/mol x 0.0084 mol = 2.3352 g
To know more about PbCl₂ refer to: https://brainly.com/question/9581816
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