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How do you determine the amount of reactant needed in this equasion?

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born2dive00:
I am trying to figure out how to calculate the amount of H2O needed, and NaOH from just the formula with the only given being the amount of Trisodium citrate.

Na3C6H5O7(aq) + 3 H2O(l) = C6H8O7(aq) + 3 NaOH(aq)
The amount in grams of the TSC is 379mg

This on line calculator says that there should be the following
https://www.webqc.org/balance.php?reaction=Na3C6H5O7%28aq%29%2BH2O%28l%29%3DC6H8O7%28aq%29%2BNaOH%28aq%29

Na3C6H5O7 = 1 coefficient, with a molar mass of 258.06900784,   1.4685994384687058 Moles, and a weight of 379 g
H2O = 3 coefficient, with a molar mass of 18.01528, 4.405798315406117 moles, and a weight of 79.37169027556952 g

which equals the following

C6H8O7(aq) = 1 coefficient, with a molar mass of 192.12352,    1.4685994384687058 moles, and a weight of 282.15249358863116 g
NaOH(aq) = 3 coefficient, with a molar mass of 39.99710928,    4.405798315406117 moles and a weight of 176.21919668693837 g

So I know to get the # of moles you divide the weight by the AMU which in the case of Na3C6H507 would be 379/258... which will give us the 1.46 moles.

Now I also know that the Na3C6H5O7 and the C6H8O7 moles will be the same as it is a tri for tri replacement. So if i take 1.468599 X 192.12352 i get  282.15249358863116 g of citrate
But what I can not work out is how to calculate the amount of moles of water and then to get weight in grams when the amount of moles are not known. (or at least i cant figure it out)

can some one please show me step by step how to calculate the moles of water, and or citric acid. I am obviously doing something wrong as I can not get it to calculate correctly.

Thank you guys.

chenbeier:
The reaction dont takes place. Trisodiumcitrate will only dissolved in water and get dissociation.

1 mol Na3 citr. need 3 mol H2O

m(Na3Citr)/M(Na3Citr) = m(H2O)/3*M(H2O)

born2dive00:
Ok if i take 3 moles of water which was my original thought, and mutiply it by 18.01528 i get 54.04584g of water, the calculator says I should be using 4.405 moles of water at 79.37g

So what am i doing wrong.
also If I use only 3 moles of NaOH i get 119.9g which through another method I know is wrong. I Know through another verification method, i should have 176.21g of NaOh.
So I know the calculator is correct when it states 4.40 moles

but what am i doing wrong that it is not coming out correctly?



--- Quote from: chenbeier on August 28, 2019, 02:19:41 PM ---The reaction dont takes place. Trisodiumcitrate will only dissolved in water and get dissociation.

1 mol Na3 citr. need 3 mol H2O

m(Na3Citr)/M(Na3Citr) = m(H2O)/3*M(H2O)

--- End quote ---

chenbeier:
You have 379 mg trisodiumcitrate this correspond to 79,3 mg water. Nothing more. What do you expect.

Borek:
You don't get three moles of water - that would be for 1 mole of citrate. Coefficients tell you in what ratio things appear in the reaction, not what are absolute amounts. You have to scale everything up/down to the amount of the substance given.

That first, then the reaction doesn't make much sense and you are mistaking mg for g.

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