The question is: All fatty acids react with NaOH in a 1:1 mole ratio (this is a typical acid- base reaction). It is known that 125 mg of lauric acid is completely neutralized by 64.5 mg of NaOH. using this information and your answer to problem 5, determine the molecular formula of lauric acid.
Okay, so number 5 was: what mass is hydrogen will react with 15.00 grams of tungsten (VI) oxide, according to the following equation? WO3 (s) + 3 H2 (g) ==> W (s) + 3 H2O (l)
And for that problem (5) I got the answer to be .1957g H2
2007-08-16 18:38:20 · 3 answers · asked by Melody 2 in Science & Mathematics ➔ Chemistry
.1957 grams of dihydrogen?
2007-08-16 18:46:42 · answer #1 · answered by Zachary 3 · 0⤊ 0⤋
I think there is something wrong with this problem. If the reaction is really one to one, then since the molecular mass of lauric acid 200.32 g/mol then the amount of NaOH needed to neutralize 125 mg lauric acid would be:
125mg LA/ (200.32g/mol LA)* (40.0g NaOH/mol LA) = 24.96 mg NaOH
LA here stands for lauric acid. We can see that what was needed to neutralize 125 mg LA was 24.96mg. Was the 64.5 mg NaOH pure NaOH or does it also contain the mass of the solvent if it were in a solution?
2007-08-16 19:14:22 · answer #2 · answered by Aken 3 · 0⤊ 0⤋
The molar mass of NaOH is 40 (23+1+16)
If the reaction is 1 to 1 so the mass of products is proportional to molecular mass and
mass of lauric acid (125/64.5)*40 =78
the atomic mass of WO3 is 184+48=232.
so 15g = 15/232=0.0647mole of W. You need 3 moles of H2
for 1 mole of WO3. So 0.194 mole of H2 with weight 2
so you need 0.388G of H2
2007-08-16 18:55:01 · answer #3 · answered by maussy 7 · 0⤊ 0⤋