Consider the reaction:
2 CH3OH(g) + 3 O2( g)¡2 CO2( g) + 4 H2O(g)...

Question

Consider the reaction: 2 CH3OH(g) + 3 O2( g)¡2 CO2( g) + 4 H2O(g) Each of the molecular diagrams represents an initial mixture of the reactants. How many CO2 molecules form from the reaction mixture that produces the greatest amount of products?

Accepted Answer

Welcome back everyone in this example, we're told that the following diagrams are molecular representations of the initial amount of the reactant is for the reaction below. So we have two moles of metal alcohol reacting with three moles of oxygen gas to produce two moles of carbon dioxide and four moles of water. Were given the legend below where we have this molecule representing our mental alcohol and then the two pink atoms bonded to one another representing our oxygen gas. And then we need to calculate the number of water molecules produced from the mixture that yields the highest amount of products. So we want to figure out which of these molecular depictions, A, B, C or D will produce the highest amount of water molecules as a product. So, beginning with our solution for molecular diagram A which will solve for over here, on the right, we want to count the amount of initial methyl alcohol first in this depiction. So we have one mole, two moles, three moles formals and five moles of methyl alcohol. So we'll write that down of five moles of ch, 30. H that are present initially. We want to multiply this by our number of water molecules that should be produced according to our given reaction. And that is four moles of water molecules. So we'll multiply by four moles of water and we want to divide by the amount of methyl alcohol reacting according to our reaction and according to our reaction, that is two moles of methyl alcohol. So we'll circle this in red and divide by that value. two moles of methyl alcohol actually reacting. And so this is going to give us a value of our water molecules that should actually be produced. And this is equal to 10 moles of water produced from methyl alcohol in depiction A. Moving on to our second reactant because we recognize we also have water. Or sorry, oxygen gas in molecular diagram A. We can count 123456 moles of our oxygen gas present initially. So we have six moles Of 02 gas initially. We want to multiply this also by the amount of water that is produced according to our reaction. That is formals of water. And we're going to divide by our oxygen gas that is actually reacting according to our reaction, and that is three moles of oxygen gas that will circle that in purple. So we'll divide by three moles of 02 gas that are actually reacting according to our reaction. And this is going to give us a value equal to eight moles of water that is produced. So comparing these two amounts of moles of water produced, we would say, since this is the lower amount oxygen gasses are limiting reactant. And so this is going to be the actual amount of water that is produced from molecular diagram A. So we'll label this here as limiting. So again, specifically the oxygen gas is the limiting reactant, since it produced the lowest amount of water. So, moving on to molecular diagram B, we want to see if B will produce a higher amount of water from the limiting reactant. So beginning with our metal alcohol, again, we have 123 moles of methyl alcohol. Initially we're going to multiply by the formals of water that should be produced according to our given reaction in the prompt. And let's make this division sign better in our denominator. We want to divide by the number of methyl alcohol that is actually reacting and as we stated above, that is two moles of methyl alcohol from our reaction actually reacting. This is from our prompt. And so what we're going to get here is a value equal to six moles of water that is produced from methyl alcohol in molecular diagram p. So moving on to oxygen gas, because we also have that reacting, We would count 123456789 moles of our oxygen gas that are initially present. We're gonna multiply this by our four moles of water that should be produced according to our given reaction in the prompt, and then we're dividing by our number of oxygen gas that is actually reacting according to the reaction given in our prompt and above that is three moles of oxygen gas. So we divide again by three moles of 02 gas. And what we're going to get here is a value equal to 12 moles of water that is produced. And so comparing these two moles of water produced in molecular diagram B. We would say. Since we have just six moles of water produced from Method alcohol in molecular diagram B. Method alcohol is our limiting reactant. So we'll label that here, and this is going to be the actual amount six moles of water that is produced from molecular diagram B. So moving on to compare with choice C. Or sorry, molecular diagram. See, we'll do the work for that below here, we can fit it. So beginning with our metal alcohol, we can count 1234567 moles of methyl alcohol that are initially present. We want to multiply this by the moles of water that should be produced according to our reaction. That's four moles of water as we stated above from the prompt we're dividing by our moles of methyl alcohol that are actually reacting and that is just the two moles of metal alcohol as we followed before from our given reaction that are actually reacting. So this is going to give us a amount of water actually produced equal to 14 moles of water that is produced from ethyl alcohol. And so, following the same steps for our second reactant for molecular diagram, see, we can count for oxygen gas 1234 moles of oxygen gas for molecular diagrams, see, so we say, four moles of 02 gas and sorry about that Four moles of 02 gas that are initially present, multiplied by the formals of water that we should produce according to our given reaction, we're going to divide by the amount of oxygen gas that is actually reacting from our given reaction. In the prompt, that is three moles of oxygen gas actually reacting. And what we're going to get here is a value equal to 5.33 moles of water That is produced from oxygen gas and molecular diagram. See and so comparing 14 moles of water produced to 5.33 moles of water produced, we would say since 5.33 is less than 14 moles, we would say that this is going to be the actual amount of water produced from oxygen gas. So oxygen gasses are limiting reactant. And so so far, we wanna see out of these three moles of water produced by our limiting reactant in molecular diagrams, A, B and C, which one of these is the highest amount of moles of water produced. Because we want to find that value to answer this question, according to our prompt. And so we would say that since we have eight moles of water produced from oxygen gas in part A that's a pretty high value compared to six. And then compared to 5.33 in B and C. And so right now we can already rule out choices B and C, because right now eight is definitely the highest value of moles of water produced from the limiting reactant. So now all we have to do is compare D to A to see if D will compete. And so we'll just do the work for D. Over here. So beginning with our moles of methyl alcohol, we have one mole of methyl alcohol initially present in molecular diagram D. We're going to multiply this by the moles of water produced according to our reaction. That's four moles of water. And we're going to divide by our actual amount of methyl alcohol reacting according to our reaction in the prompt. And that is the two moles of methyl alcohol actually reacting from our prompt. So we divide by two moles of methyl alcohol And what we're going to get here is a value equal to 22 moles of water that is produced from ethyl alcohol in molecular diagram D. And then moving on to our second reactant, we have two moles of oxygen gas. So we have two moles of 02 gas initially present in molecular diagram D. Multiplied by the formals of water that should be produced according to our given reaction in the prompt and divided by our denominator. Where we have the amount of oxygen gas actually reacting according to the prompt for a given reaction. That is three moles of oxygen gas actually reacting. What we're going to get here is a value equal to 2.67 moles. I'm sorry moles of water that is produced. We would say that too is less than 2.67 and so are limiting reactant is going to be the methyl alcohol which produces the two moles of water here. So this is our actual amount of water that is produced. And so because we already rolled out molecular diagrams B and C, comparing the two moles of water produced from molecular diagram D to the eight moles of water produced from molecular diagram A. We would say that too is definitely less than eight. And so this is not a high amount of water produced. We want the highest amount of water as a product produced from these molecular diagrams. So ruling out choice D. That leaves us with a molecular diagram a as the best diagram that showed the highest amount of moles of our product, water that is produced. And so this value here where we have these eight moles of water produced is going to be our final answer to complete this example. So, I hope that everything that I went through is clear. If you have any questions, please leave them down below and I will see everyone in the next practice video

Consider the reaction: 2 CH3OH(g) + 3 O2( g)¡2 CO2( g) + 4 H2O(g) Each of the molecular diagrams represents an initial mixture of the reactants. How many CO2 molecules form from the reaction mixture that produces the greatest amount of products?

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