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CH3NH2·HCl (67.52g) + NaOH (40g) → CH3NH2 (31.06g) + NaCl (58.44g) + H2O (18.01g)
200ml of 40% aqueous methylamine => 80g CH3NH2 dissolved in 200g H2O
If I use (174g of methylamine hcl + 67% aqueous NAOH solution)
CH3NH2.HCL 174g
NAOH 103g + H20 154g
I obtain (200ml of 40% aqueous methylamine):
CH3NH2 80g
NACL 150g
H20 200g
Would it be possible?
200ml of 40% aqueous methylamine => 80g CH3NH2 dissolved in 200g H2O
If I use (174g of methylamine hcl + 67% aqueous NAOH solution)
CH3NH2.HCL 174g
NAOH 103g + H20 154g
I obtain (200ml of 40% aqueous methylamine):
CH3NH2 80g
NACL 150g
H20 200g
Would it be possible?
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REAGENTS
CH3NH2·HCl 67,518 g/mol
NaOH 40 g/mol
PRODUCTS
CH3NH2 31.0571 g/mol
NaCl 58.4428 g/mol
H20 18 g/mol
I decided to do some calculations myself to clarify the matter for myself; maybe it will clarify for others.
Methylamine aq 40% → 1 CH3NH2 (31g) + 2.5 H2O (45g)
31g divided by 31g + 45g = 40%
The density of Methylamine 40% (w/w) in water is 0.897 g/mL, that is, 200ml has 179.4 grams.
40% methylamine = 71.76g (2.31 mol)
60% water = 107.64g
I have seen recommendations to use a 1:5 ratio of NaOH to H2O and add dropwise to an ice bath aqueous solution of methylamine.hcl. Perhaps the use of a lower concentration of NaOH for basification has the objective of making the reaction more peaceful, not heating, not releasing methylamine in gas; but I can't say for sure.
To clarify, the 1:5 ratio of NaOH to H2O refers to a 31% concentration.
1 mol NaOH => 40g
5 mol H20 => 90g
40g divided by 40g + 90g = 31%
Methylamine.hcl has a water solubility of 1g/mL (at 20 °C)
1 mol CH3NH2.HCL => 67.52g would need 4 mols of H20 (72g/72ml) to completely dissolve, a concentration of 48%:
67.52g divided by 67.52g + 72g = 48%
The balanced reaction equation is:
CH3NH2·HCl + NaOH → CH3NH2 + NaCl + H2O
The equation involving solutions of methylamine.hcl with NaOH solution above would look like this:
(1 CH3NH2·HCl + 4 H2O) (48% aq) + (1 NaOH + 5 H2O) (31% aq) → 1 CH3NH2 + 1 NaCl + 10 H2O
To obtain 2.31 mol of methylamine equivalent to 200ml of Methylamine 40% (w/w) in water, we would have:
(2.31 CH3NH2·HCl + 9.24 H2O) (48% aq) + (2.31 NaOH + 11.55 H2O) (31% aq) → 2.31 CH3NH2 + 2.31 NaCl + 23.1 H2O
In grams, we would have (rounded):
(156g CH3NH2·HCl + 166g H2O) + (93g NaOH + 208g H2O) → 72g CH3NH2 + 135g NaCl + 416g H2O
We would have 72g of methylamine with 135g of NaCl dissolved in 416g (ml) of H20.
An aqueous solution of 33% methylamine + NaCl, being 11% methylamine and 22% NaCl.
72g + 135g divided by 72g + 135g + 416g = 33%
72g/623g (11%) + 135g/623g (22%) = 33%
I don't know if excess water would disturb the synthesis of this topic, but just removing excess water with a drying agent without removing NaCl would not solve it, as we would have solubility problems (maximum saturation). I couldn't tell if the NaCl would precipitate or if the methylamine would be released as a gas, or perhaps both.
CH3NH2·HCl 67,518 g/mol
NaOH 40 g/mol
PRODUCTS
CH3NH2 31.0571 g/mol
NaCl 58.4428 g/mol
H20 18 g/mol
I decided to do some calculations myself to clarify the matter for myself; maybe it will clarify for others.
Methylamine aq 40% → 1 CH3NH2 (31g) + 2.5 H2O (45g)
31g divided by 31g + 45g = 40%
The density of Methylamine 40% (w/w) in water is 0.897 g/mL, that is, 200ml has 179.4 grams.
40% methylamine = 71.76g (2.31 mol)
60% water = 107.64g
I have seen recommendations to use a 1:5 ratio of NaOH to H2O and add dropwise to an ice bath aqueous solution of methylamine.hcl. Perhaps the use of a lower concentration of NaOH for basification has the objective of making the reaction more peaceful, not heating, not releasing methylamine in gas; but I can't say for sure.
To clarify, the 1:5 ratio of NaOH to H2O refers to a 31% concentration.
1 mol NaOH => 40g
5 mol H20 => 90g
40g divided by 40g + 90g = 31%
Methylamine.hcl has a water solubility of 1g/mL (at 20 °C)
1 mol CH3NH2.HCL => 67.52g would need 4 mols of H20 (72g/72ml) to completely dissolve, a concentration of 48%:
67.52g divided by 67.52g + 72g = 48%
The balanced reaction equation is:
CH3NH2·HCl + NaOH → CH3NH2 + NaCl + H2O
The equation involving solutions of methylamine.hcl with NaOH solution above would look like this:
(1 CH3NH2·HCl + 4 H2O) (48% aq) + (1 NaOH + 5 H2O) (31% aq) → 1 CH3NH2 + 1 NaCl + 10 H2O
To obtain 2.31 mol of methylamine equivalent to 200ml of Methylamine 40% (w/w) in water, we would have:
(2.31 CH3NH2·HCl + 9.24 H2O) (48% aq) + (2.31 NaOH + 11.55 H2O) (31% aq) → 2.31 CH3NH2 + 2.31 NaCl + 23.1 H2O
In grams, we would have (rounded):
(156g CH3NH2·HCl + 166g H2O) + (93g NaOH + 208g H2O) → 72g CH3NH2 + 135g NaCl + 416g H2O
We would have 72g of methylamine with 135g of NaCl dissolved in 416g (ml) of H20.
An aqueous solution of 33% methylamine + NaCl, being 11% methylamine and 22% NaCl.
72g + 135g divided by 72g + 135g + 416g = 33%
72g/623g (11%) + 135g/623g (22%) = 33%
I don't know if excess water would disturb the synthesis of this topic, but just removing excess water with a drying agent without removing NaCl would not solve it, as we would have solubility problems (maximum saturation). I couldn't tell if the NaCl would precipitate or if the methylamine would be released as a gas, or perhaps both.
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