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Topic: {stoichiometry}  (Read 2783 times)

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Offline jldz140

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{stoichiometry}
« on: March 31, 2014, 07:53:59 PM »
2 NaOH + CuSO4 --> Na2SO4 + Cu(OH)2

100g of Cu(OH)2 is needed.  Assuming 100% yield, how much of 1M NaOH and 0.5M CuSO4 must be used in mL?

Any help would be greatly appreciated!
« Last Edit: March 31, 2014, 08:48:07 PM by Arkcon »
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Offline Arkcon

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Re: {stoichiometry}
« Reply #1 on: March 31, 2014, 08:51:01 PM »
You have grams of product needed, and you have mL of reactants at a certain concentration.  You have to equate the two.  Fortunately, you have a chemical equation to relate the two.  What are the units of the chemical equation?  ANd can you begin to convert them?
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Hey, I'm not judging.  I just like to shoot straight.  I'm a man of science.

Offline jldz140

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Re: {stoichiometry}
« Reply #2 on: March 31, 2014, 08:55:34 PM »
I'm trying to follow you. But--I really don't know where to start. I don't even know where to begin.
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Offline critzz

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Re: {stoichiometry}
« Reply #3 on: March 31, 2014, 09:33:22 PM »
From the equation you can see that 2 moles of NaOH and 1 mole of CuSO4 give 1 mole of Cu(OH)2.

Start by converting the 100g of Cu(OH)2 into moles.

Using the equation, you can then calculate the amount moles needed of the NaOH and CuSO4.
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