Chemistry, asked by bananapeels, 2 months ago

3. What mass of lead(II) sulphate would be produced by the action of excess dilute sulphuric acidon 10 g of lead nitrate dissolved in water

Answers

Answered by Havyasdore

1

Explanation:

Assuming that we have two reactants A and B.

Reactant A is 100 % finished after complete reaction while reactant B is still left after the reaction.

Reactant A is a limiting reactant in this case.

Limiting reactant will always decide the yield of the product obtained.

hope it helps you

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