Determining Precipitate Yield (g) in Aqueous Solution

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Given Question: Given the reaction: K3(PO4)+ Ni(SO4)→ K3(SO4)(aq)+Ni(PO4)(s), with 100.0 ml of K3PO4 and 200.0 ml of NiSO4 under standard conditions, how many grams of precipitate form? 

Step 1: Determine the limiting reactant and the precipitate. 

In order to determine the limiting reactant, one must convert all reactants to moles (mol). In this case, we can use the molarity equation (Molarity=moles/Liter) to find moles of Potassium phosphate and Nickel Sulfate. The reactant with the least amount of moles is our limiting reactant.

In this case, potassium phosphate is our limiting reactant (.02 moles).

We can also determine the precipitate by observing which solid is formed in the reaction. We know precipitates in this case are usually white, crystalline solids that form in aqueous solutions, so it’s safe to say the nickel phosphate is our precipitate.

Step 2: 

We set up a basic dimensional analysis equation to find how many moles of nickel phosphate are made from .02 moles of potassium phosphate. Remember your units, and always make sure they cross out! You should have a yield of 3.072 moles of nickel phosphate. Feel free to check numbers!


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(Source: OSU chemistry deptarment, 2015)




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