N-Methylacetamide Hydrolysis

[CryoThio]

I'm trying to do the math on the hydrolysis of N-MethylAcetamide (abbriviating it to NMA) in order to get the methylamine.

I would be using basic hydrolysis, so there will be sodium acetate and methylamine in solution. The sodium acetate likely will not interfere with a mercury amalgam reduction of various ketones, so I'm not concerned about removal.

I want to know how much of each reagent to react so as to get a 40% solution of methylamine.

Say one has 100 grams of NMA; how much water and sodium hydroxide will be needed to produce a 40% methylamine solution?

 

[loadingST]

Learn abaut mollar masses(every chem have its own mass) so to make equal you need to check the masses and use 1:1molar mass

 

[CryoThio]

Learn abaut mollar masses(every chem have its own mass) so to make equal you need to check the masses and use 1:1molar mass

loadingSTSo would I calculate the molar masses of each chemical, water, sodium hydroxide, and the NMA, and use a 1 to 1 ratio of these?

 

[CryoThio]

So would I calculate the molar masses of each chemical, water, sodium hydroxide, and the NMA, and use a 1 to 1 ratio of these?

CryoThioJust trying to make sure I understand

 

[CryoThio]

So would I calculate the molar masses of each chemical, water, sodium hydroxide, and the NMA, and use a 1 to 1 ratio of these?

CryoThioAs in "use whater weight of each chemical to make the molar mass a one to one ratio?

 

[loadingST]

Nmethylacetamide 73,09g one moll, naoh 39,9g, disolve naoh in water and add slowly to your Acetamide...

 

[CryoThio]

Nmethylacetamide 73,09g one moll, naoh 39,9g, disolve naoh in water and add slowly to your Acetamide...

loadingSTI understand completely, and I assume one mole of water as well? Should be 18 grams I think. Or is more needed.