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What quantity in moles of precipitate will be formed when 100.0 mL of 0.150 M LiBr is reacted with excess Pb(NO₃)₂ in the following chemical reaction?
2 LiBr(aq) + Pb(NO₃)₂(aq) → PbBr₂(s) + 2 LiNO₃(aq)

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Answer : The number of moles of precipitate,  formed will be 0.01 moles.

Explanation : Given,

Concentration of NaBr = 0.200 M

Volume of solution = 100.0 mL = 0.1 L      (1 L = 1000 mL)

First we have to calculate the moles of NaBr.

Now we have to calculate the moles of precipitate,  formed.

The balanced chemical reaction is:

From the balanced chemical reaction we conclude that:

As, 2 moles of NaBr react to give 1 mole of

So, 0.02 moles of NaBr react to give  mole of

Therefore, the number of moles of precipitate,  formed will be 0.01 moles.

Explanation:

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