Chemical Forums
Chemistry Forums for Students => Analytical Chemistry Forum => Topic started by: kiloa on April 17, 2019, 03:32:39 PM
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dear all,
if you need 70g sodium molybdate for preparing certain reagent but i have ammonium molybdate instead of sodium molybdate ....how can i calculate the required weight of ammonium molybdate ?
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What do you need it for? It might be that you can't easily swap as you suggest.
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i used one of them to precipitate phosphorus as Phosphomolybdate ,so the molybdate is required component
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any help ??!!
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Hint: use equimolar amount.
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what do you mean ?
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No matter which salt you use, you need the same number of moles.
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yes yes , i have known what you do mean ... thanks so much