CHRIS25
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Please check my calculations?
Hi, I have to mix 45g of sodium bisulphite into 500ml water. I only have sodium metabisulphite.
I took the g/moles of both substances: Based on the knowledge that metabisulphite becomes bisulphite automatically when dissolved I concluded the
following logic????
45g x 104g/mol = 4680; then I divided 4680 by 190g/mol for the metabisulphite = 24.6 grams. Therefor I use 24.6 grams of bisulphite to equal the
original 45 grams of sulphite required?
Have I done this correctly or am I way off the beaten track?
Regards, Chris.
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barley81
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Unfortunately that's not correct.
Here's how I would do it:
45g/(104g/mol) = 0.43 mol. You need 0.43 moles of sodium bisulfite. Since one mole of sodium metabisulfite converts into two moles of sodium bisulfite
in water, you need 0.22 moles of sodium metabisulfite. 0.22 mol * (190g/mol) = 41.1 grams of sodium metabisulfite.
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CHRIS25
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Hi Barley, it seems to me that I would never have worked that one out because even when I look at the formulas for both chemicals I would never have
known that I need to convert metabisulphite into two moles of sodium bisulphite in water. The rest of your explanation I can see what I did wrong.
Thankyou. Many appreciations.
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