Single Status Update
you walked right by me. :P I would have said hi but you were on the phone. Plus, I was painfully engrossed in the following chemistry problems. I have spent upwards of 4 hours on the following types of chemical problems and have failed to get more than 2 in a row correct. I only recently now understand how to do the problems, but I still struggle with the steps.
You have two reagents. You combine them and get a product. Applying simple chemical math, you can figure out which one is the limiting reagent and what the maximum amount of the product that can be formed from the reagent.
An example problem:
For the following reaction, .0327 moles of barium hydroxide are mixed with 2.84 grams of sulfuric acid.
barium hydroxide (aq) + sulfuric acid (aq) results in the products: barium sulfate (aq) + water (l)
What is the limiting reagent?
What is the maximum amount of water that can be found in grams?
First you convert 2.84 grams of sulfuric acid to moles. To do this, divide 2.84 grams by the number of grams that make up 1 mole of sulfuric acid. We know that Sulfuric Acid is H_2_SO_4_. Using the periodic table we can take the sum of the constituent elements' mass to get the molar mass of H2SO4: 98.0794grams. 2.84g/98.0794g = .02896 moles. Next we balance the equation. First we use google to convert those fancy compounds into chemical formulas and we get the following formula that we can now balance.
Ba(OH)_2_ + H_2_SO_4_ = BaSO_4_ + H_2_O becomes
Ba(OH)_2_ + H_2_SO_4_ = BaSO_4_ + 2H_2_O
First, count up the number of each element on each side of the equation.
On the left we have 1Ba, 6O, 4H, 1S
On the right we have 1Ba, 5O, 2H, 1S
If we put a 2 in front of the H_2_O that should do it. We'll now have 4H on the right and 6O on the right.
Now we have setup this problem to find the limiting reagent.
We have .02896 moles of H2SO4 and .0327 moles of Ba(OH)_2_
Now we compare the two to see which one is less moles after going through the reaction. First we divide the number of moles of H_2_O, which we found out to be 2, by the number of moles of one of our reagents and then multiply by the number of moles of the reagent. H_2_O is what we are being asked will produce the max number of grams, so it goes in the numerator. We have 1 mole of both our reagents, so we know the number we get for both reagents will be proportionally the same. Therefore we don't need to do the simple math on both of them. We know that .02896 moles of H2SO4 is proportionally less than .0327 moles of Ba(OH)_2_. So we multiply .02896 by 2 to get the number of moles of our limiting reagent: .0579 moles. We are not quite done, though, since the problem asks us for the maximum amount of water in grams and we only know the maximum amount of water in moles. To convert to grams we simply multiply the number of grams that make up 1 mole of water, which is 18.0152 grams. We get 1.043 grams of water. I guess it's not that simple. But the maths for the problem is very straightforward. It's just tricky knowing what and when to multiply and divide and why.
Check this out.