How Much Precipitate Forms in a Copper Sulfate and Sodium Hydroxide Reaction?

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The discussion centers on calculating the mass of precipitate formed in a reaction between copper(II) sulfate and sodium hydroxide. A solution of copper(II) sulfate is prepared, and when mixed with sodium hydroxide, a precipitate of copper(II) hydroxide (Cu(OH)2) forms. The initial calculation yielded a mass of 0.762g for the precipitate, but there is uncertainty about whether the correct limiting reagent was identified, as both sodium hydroxide and copper sulfate could limit the reaction. Additionally, the ambiguity regarding the form of copper sulfate—whether it is pentahydrate or anhydrous—adds complexity to the calculation. Clarifying these details is essential for an accurate determination of the precipitate mass.
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Homework Statement


A solution of copper 2 sulphate is prepared by placing 5.80g of copper 2 sulphate in a flask and adding enough water to prepare 125 ml of solution. When a 25.0ml sample of this solution is added to 125ml of 0.125 mol/L sodium hydroxide a precipitate forms. Calculate the mass of the precipitate formed in the reaction.

Homework Equations


n=cv
m=nM

The Attempt at a Solution


The percipitate i found would be cu(OH)2 which would have the same mole as NaOH (n=cv) and then u would use the formula m=nM to find mass. The answer i got is 0.762g but i didnt use any info about the copper 2 sulphate, is this right?
 
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You have to find what is the limiting reagent - it can be NaOH, it can be copper sulfate.

I find the question ambiguous - it doesn't say whether it is a copper sulfate pentahydrate (the most common form) or the anhydrous form.
 
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