Chemistry - Stoichiometry
Comments
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Alright, so I'm assuming that there is an excess of sulfuric acid so if thats the case this is how it will go down.
You first want to know the number of mols that 200g of sodium hydroxide has by dividing by it's molar mass.
That gets you 200g divided by 40g/mol which = 5 mols.
You then have to look at the mol ratios by looking at the balanced equation. You see that for every Sodium Sulfate there is 2 Sodium Hydroxide required. That means that there is one half of the number of mols because it is a 2:1 ratio.
So you take that 5 mols and divide by 2 (or multiply by 1/2). That gets you 2.5 mols.
Now that you have the mols of Sodium Sulfate you want to multiply by the molar mass to finally get you the grams.
So 2.5mols x 142 g/mol = 355 grams. -
Maybe writing down the units while you do the calculation might help you understand it.
So,
200 g NaOH x ( 1 mol NaOH / 40 g NaOH) = 5 mol NaOH
5 mol NaOH x ( 1 mol Na2SO4 / 2 mol NaOH) = 2.5 mol Na2SO4
2.5 mol Na2SO4 x ( 142 g Na2SO4 / 1 mol Na2SO4) = 355 g Na2SO4
If you write the units down, you will understand what's happening instead of just playing with the numbers. Also, this will get rid of the errors or any confusions as to where to put the numbers in terms of denominator or numerator since all units will cancel out except for the answer you are looking for if you have wrote them correctly. -
I really didn't like stoich also. but if you just remember to label as you convert from grams to moles, then you should be okay. If you are having trouble with the nomenclature part. just go look up "Golden sheet of nomenclature" and memorize/look for patterns for them. and lol, this has nothing to do with Cf.
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