For each of the following unbalanced chemical equations, suppose 10.0 mg of each reactant is tak

For each of the following unbalanced chemical equations,...

Key Concepts

Theoretical Yield

The theoretical yield is the maximum amount of product that can be formed from the limiting reactant if the reaction proceeds perfectly. It is calculated using the stoichiometric relationships and is used to evaluate the efficiency of chemical reactions.

Limiting Reactant

The limiting reactant is the substance that is completely consumed first according to the stoichiometric ratios, thereby limiting the amount of product formed. Identifying the limiting reactant is fundamental for calculating the theoretical yield of a reaction.

Balanced Chemical Equations

Balanced chemical equations are essential because they reflect the conservation of mass in a reaction. They provide the stoichiometric ratios between reactants and products, which are crucial for determining how much of each substance is involved in the reaction.

Molar Mass and Mole-to-Mass Conversions

Molar mass is the mass of one mole of a substance and is used to convert between grams (or milligrams) and moles. This conversion is vital in stoichiometric calculations to relate the measured mass of a substance to the number of particles or moles present.

Stoichiometry

Stoichiometry involves using the mole ratios from the balanced chemical equation to calculate the amounts of reactants and products. It allows one to determine how much product can be formed from given quantities of reactants and to scale the reaction accordingly.

Transcript

00:01 This calculation, similar to the previous calculations, have four unbalanced chemical reactions.

00:07 The first thing you need to do is balance the chemical reactions.

00:11 Then, assuming that we are starting with 10 milligrams of each of the reactants, figure out which is the limiting reactant at that amount.

00:20 Once you've identified the limiting reactant, then use the amount of the limiting reactant to calculate the amount of product produced, amount of each product produced.

00:35 So let's start with this balanced, or this unbalanced chemical reaction, carbon monoxide with hydrogen gas produces methanol.

00:44 To balance it, all we need to do is nothing.

00:48 It's already, actually, we need to put a two right here, which i'll do in just a second.

00:55 So if we have 10 milligrams of each reactant, we'll start with 10 milligrams or 0 .0 grams of carbon monoxide.

01:05 Divide by its molar mass to get moles carbon monoxide, and then go from moles carbon monoxide to moles of, there's only one product here, methanol, using the one -to -one stoichiometry.

01:20 And we get 3 .57 times 10 to the negative 4 moles of methanol.

01:26 If we have 10 milligrams or 0 .01 grams of hydrogen gas, we can convert the grams to moles by dividing by the molar mass of h2, and then go from moles of h2.

01:38 To moles of methanol using the stoichiometry.

01:43 And we get 2 .48 times 10 to negative 3 moles of the product.

01:48 So because carbon monoxide produces the least amount of product, it is the limiting reactant.

01:54 And all we need to do is carry out this calculation one step further.

01:59 Once we have moles of methanol, we'll multiply by the molar mass of methanol to get the grams of methanol.

02:06 0 .114 grams, methanol.

02:12 For the next one, we have aluminum, reacting with iodine, producing aluminum iodide.

02:25 To balance this, we have three iodines and two iodine, so we don't need to introduce a fraction.

02:33 I'm going to put a two here, so we have six iodines.

02:35 That gives us two aluminums.

02:38 So we'll put a two on the aluminum and a three on the iodine, the i2.

02:46 Then if we have 10 milligrams, which is 0 .01 grams of aluminum, we'll convert it to moles aluminum by dividing by the molar mass, and then go from moles of aluminum to moles of the only product, aluminum iodide, a 2 -2 -2 relationship.

03:02 And we get 3 .71 times 10 of the negative 4 moles, aluminum iodide.

03:07 We'll then start with 10 milligrams or 0 .01 grams of iodine, divide by its molar mass to get moles iodine, and then go from moles iodine 3 to moles aluminum iodide, using the stoichiometry, and we get 2 .63 times 10 to negative 5 moles of aluminum iodide, because this is the least amount of product.

03:30 This is our limiting reactant.

03:33 And all we need to do to get the mass of the only product, aluminum iodide, is take this calculation one step further and multiply by the molar mass of aluminum iodide, which you can calculate or look up, point six nine grams and you get point zero one zero seven grams your book may have these as milligrams to convert grams to milligrams you just multiply by a thousand so if we multiply by a thousand one two three it'd be 11 .4 milligrams and down here it would be 10 .7 milligrams okay for the next one we have calcium oxide reacting with hydrobromic acid, producing calcium bromide, and water.

04:36 Let's back up.

04:38 To balance this one, we have two bromines.

04:42 We're going to have to put a two here.

04:46 Now that gives us a total of one, two, three, four hydrogens.

04:51 We've got to put a two there to get four hydrogens.

04:54 Now we've got two oxygens, but we've got two oxygens and one calcium and one calcium.

05:00 So it's balanced.

05:05 Now let's start with 10 milligram.

05:07 Of calcium hydroxide or 0 .010 grams, convert the grams to moles by dividing by the molar mass of calcium hydroxide, then use the stoichiometry to calculate the moles of any product.

05:24 It doesn't matter which product you choose.

05:28 The limiting reactant will always produce the least amount of both products...

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