Methanol, CH3 OH, is a clean-burning, easily handled fuel....

Methanol, $\mathrm{CH}_{3} \mathrm{OH}$, is a clean-burning, easily handled fuel. It can be made by the direct reaction of $\mathrm{CO}$ and $\mathrm{H}_{2}$. $$ \mathrm{CO}(\mathrm{g})+2 \mathrm{H}_{2}(\mathrm{~g}) \longrightarrow \mathrm{CH}_{3} \mathrm{OH}(\ell) $$ (a) Starting with a mixture of $12.0 \mathrm{~g} \mathrm{H}_{2}$ and $74.5 \mathrm{~g} \mathrm{CO}$, which is the limiting reactant? (b) What mass of the excess reactant, in grams, is left after reaction is complete? (c) What mass of methanol can be obtained, in theory?

Methanol, $\mathrm{CH}_{3} \mathrm{OH}$, is a clean-burning, easily handled fuel. It can be made by the direct reaction of $\mathrm{CO}$ and $\mathrm{H}_{2}$.
$$
\mathrm{CO}(\mathrm{g})+2 \mathrm{H}_{2}(\mathrm{~g}) \longrightarrow \mathrm{CH}_{3} \mathrm{OH}(\ell)
$$
(a) Starting with a mixture of $12.0 \mathrm{~g} \mathrm{H}_{2}$ and $74.5 \mathrm{~g} \mathrm{CO}$, which is the limiting reactant?
(b) What mass of the excess reactant, in grams, is left after reaction is complete?
(c) What mass of methanol can be obtained, in theory?

Key Concepts

Stoichiometry

Stoichiometry involves the quantitative relationships between reactants and products in a chemical reaction. It relies on the balanced chemical equation to determine the proportionate amounts (in moles) required for complete conversion of reactants into products and to predict how much product will be formed from given reactant quantities.

Limiting Reactant

The limiting reactant is the substance in a chemical reaction that is completely consumed first, thereby limiting the extent of the reaction and determining the maximum amount of product that can be formed. Identifying the limiting reactant is essential for calculating the theoretical yield of the reaction.

Excess Reactant

The excess reactant is the substance that remains partially unreacted after the reaction is complete. Once the limiting reactant is entirely consumed, any remaining quantities of other reactants are considered excess. The amount of the excess reactant left can be calculated by determining the amount that reacted and subtracting it from the initial amount.

Molar Mass

Molar mass is the mass of one mole of a substance, typically expressed in grams per mole (g/mol). It serves as the conversion factor between the mass of a substance and the amount in moles, which is fundamental when using stoichiometric relationships in chemical reactions.

Theoretical Yield

The theoretical yield is the maximum amount of product that can be generated from a chemical reaction based on the amount of the limiting reactant, assuming that the reaction proceeds perfectly with complete conversion and no losses. This concept allows chemists to predict reaction efficiency and compare it with actual yields obtained in practice.

Transcript

00:01 So for the first part of this question, we are asked to find the limiting reagent for the formation of methanol.

00:09 So first what we're going to do is we are going to figure out how much methanol could be produced if we used up all of the 74 .5 grams of carbon monoxide.

00:18 Then we'll figure out how much methanol can be produced if we consume all of the 12 grams of the hydrogen gas.

00:27 And then whichever one produces the smallest amount of methanol is going to be our limiting reagent.

00:34 So first we'll start with carbon monoxide.

00:36 We have 74 .5 grams of carbon monoxide.

00:40 We can only compare carbon monoxide to methanol if they're in the terms of moles.

00:47 So first we have to convert grams of carbon monoxide into moles of carbon monoxide.

00:51 We convert grams into moles.

00:53 We just do it by taking it times its molar mass.

00:55 The molar mass of carbon dioxide is 28 grams per mole.

01:03 So for every one mole of carbon monoxide, we have 28 grams of carbon monoxide.

01:08 Now we can compare moles by looking at the coefficients in the reaction.

01:15 So we have a coefficient of one for methanol.

01:19 So that means one mole of methanol is produced.

01:23 And we have a coefficient of one for carbon monoxide.

01:28 So that means we have one mole of carbon monoxide reacting.

01:34 And that'll tell us moles of methanol.

01:37 So grams of carbon monoxide on top cancel out grams on bottom.

01:41 Moles on top cancel out moles on bottom.

01:44 74 .5 divided by 28 times 1 is going to be 2 .66 moles of methanol.

01:54 So 74 .5 grams of carbon monoxide will give us a maximum amount of 2 .66 moles of methanol.

02:02 So now let's figure it out for the hydrogen gas.

02:05 So we have 12 grams of hydrogen gas.

02:09 Again, we have to convert grams into moles by taking it times its molar mass.

02:13 So for every one mole of hydrogen gas, we have 2 grams of hydrogen gas.

02:20 Because molar mass is 2 grams per mole...

Please give Ace some feedback