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1 year ago

How many grams of pbcl2 are formed when 50. 0 ml of 0. 336 m kcl react with pb(no3)2?

Chemistry

1 answer:

Vlada [557]1 year ago

3 0

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= $\frac{0. 336 * 50}{1000}$

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: brainly.com/question/9581816

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