0.63g of lead powder were dissolved in excess nitric (V) acid to form lead (II) nitrate solution. All the lead (II) nitrate was then reacted with...

      

0.63g of lead powder were dissolved in excess nitric (V) acid to form lead (II) nitrate solution. All the lead (II) nitrate was
then reacted with sodium sulphate solution.
a) Write an ionic equation for the reaction between sodium sulphate solution and lead (II) nitrate solution.
b) Determine the mass of the lead salt formed in the reaction in (a) above
(Pb = 207, S = 32, O = 16)

  

Answers


Kavungya
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Kavungya answered the question on May 23, 2019 at 09:28


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