I wanted to add 150 mL of 18 M H2SO4 to my drain to clean it, but instead I spilled it on the floor. How much baking soda (grams of NaHCO3) do I need to pour on it to neutralize it?
H2SO4 + 2 NaHCO3 ---> Na2SO4 + 2 H2CO3
2007-01-14 18:09:35 · 4 answers · asked by justinz_1fan 2 in Science & Mathematics ➔ Chemistry
In case it wasn't clear, this is a problem for my organic chemistry class. I've been working on homework for about 14 hours now, so I'm not as lucid as I should be. :)
2007-01-14 18:26:41 · update #1
Thank you all so much! Everyone was so helpful.
2007-01-14 19:01:23 · update #2
From the reaction:
H2SO4 + 2 NaHCO3 ---> Na2SO4 + 2 H2CO3
1mole of H2SO4 requires 2 moles of NaHCO3 for neutralization.
Millimoles of H2SO4 = M1*V1 = 18*150 = 2700
Millimoles of NaHCO3 = 2700*2 =5400 for neutralization
Molecular weight of NaHCO3 = 23+1+12+48= 36 +48 = 84
Milligrams of NaHCO3 = 84*5400
Grams of NaHCO3 = 84*5.4 = 453.6 g
You need to pour 453.6 g of baking soda to neutralize the spilled the acid!
2007-01-14 18:27:35 · answer #1 · answered by Som™ 6 · 1⤊ 0⤋
H2SO4 + 2 NaHCO3 ---> Na2SO4 + 2 H2CO3
This means 1 mole H2SO4 combines with 2 moles NaHCO3.
Molecular weight of H2SO4 = 98 g/mol
Molecular weight of NaHCO3 = 84 g/mol
Therefore, 98 g H2SO4 combines with 168 g NaHCO3.
1M H2SO4 contains 98 g/1000 mL.
So, 18M H2SO4 contains 98 * 18 g/1000 mL.
Thus, 150 mL of 18M is equivalent to 98 *18 * 150 / 1000 g,
which equals 264.6 g H2SO4.
The amount of NaHCO3 needed is therefore 264.6 * 168 / 98 g
which equals 453.6 g.
2007-01-14 18:38:07 · answer #2 · answered by falzoon 7 · 1⤊ 0⤋
since you know that you have 150 ml of the acid and its 18 molar, you can find the number of moles of acid from the ratio 18M = x moles/.150 L. This tells you that you have 2.7 moles of acid. using this in conjunction with your stated equation you konw that you need 2 moles of baking soda for every mole of acid. so you will need 5.4 moles of baking soda. you can convert this to a mass by multiplying it by baking soda's molecular weight. so 5.4 moles baking soda times 84 grams/mole is 453.6 grams of baking soda
2007-01-14 18:25:38 · answer #3 · answered by andrew R 2 · 1⤊ 1⤋
... i owudl jsut pour an entire box on the floor lol.....im sure you dont have to neutrolizise it that specificly....unless its liek a word problem lol
in that case
ill be lost
2007-01-14 18:22:35 · answer #4 · answered by -Eugenious- 3 · 0⤊ 1⤋