Limiting reactant example problem 1 edited | Physical Processes | MCAT | Khan Academy

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[@D8vid]

Thank you very much sir, this video helped me lots in stoichiometry. I recently had an exam and without knowing anything about stoichiometry, I saw this video the night before the test and in the test I had all the stoichiometry problems correct.

[@Nehasharma-qo8fy]

Tq sir 😊😊

[@ebubekirak8970]

teşekkür ederim

[@ZantherStone]

I see where we get that there are more moles of H2, but where do work into the equation that we use 2mol of H2 for every mol of CO when determining what the LR is? I see the 2:1 ratio written down, but not sure how it was incorporated

[@ness5138]

5:16 He compares the moles of CO and H2 by setting it up in a ratio: (32.5 mol H2)/(12.7 mol CO). Now he can reduce that ratio to (2.56 mol H2)/(1 mol CO) by dividing the numerator and denominator by 12.7. By doing this, he can set up the ratio to be similar to the original reaction: (2 mol H2)/(1 mol CO). Notice how the original reaction and the experimental reaction both have 1 mol CO in the denominator. By comparing the two ratios, it's obvious that the experiment has an excess of H2, 2.56 mol instead of 2 mol. The way to "work in" that there is 2 mol of H2 per 1 mol of CO is to first find out the moles of every reactant. Then divide by the mole of the reactant with the least amount of moles. Remember, that a mole represents an actual number of atoms (Avogadro's Number). So by dividing the reactants by the littlest mole, we can determine the ratio of all the reactants in an experimental equation. From there, it must be compared to the original equation. The experimental number of moles will always be compared against the original equation to find the LR. There is no equation that spits out what the LR is (as far as I know).

[@Iwanttoblowmybrainsoutrn]

I'm learning about limiting reactants, theoretical yield, and mole to mole ratio and I don't understand anything and my test is in 2 days :(