Consider the following balanced chemical equation: A+5 B ⟶3 C+4 D a. Equal masses of A and B are

Consider the following balanced chemical equation: A+5 B ⟶3...

Consider the following balanced chemical equation: $$\mathrm{A}+5 \mathrm{B} \longrightarrow 3 \mathrm{C}+4 \mathrm{D}$$ a. Equal masses of A and B are reacted. Complete each of the following with either "A is the limiting reactant because _________ ”; “B is the limiting reactant because ________”; or “we cannot determine the limiting reactant because ________.” i. If the molar mass of $\mathrm{A}$ is greater than the molar mass of B, then ii. If the molar mass of $\mathrm{B}$ is greater than the molar mass of A, then b. The products of the reaction are carbon dioxide (C) and water (D). Compound A has a similar molar mass to carbon dioxide. Compound B is a diatomic molecule. Identify compound $\mathrm{B},$ and support your answer. c. Compound $A$ is a hydrocarbon that is $81.71 \%$ carbon by mass. Determine its empirical and molecular formulas.

Consider the following balanced chemical equation:
$$\mathrm{A}+5 \mathrm{B} \longrightarrow 3 \mathrm{C}+4 \mathrm{D}$$
a. Equal masses of A and B are reacted. Complete each of the following with either "A is the limiting reactant because _________ ”; “B is the limiting reactant because ________”; or “we cannot determine the limiting reactant because ________.”
i. If the molar mass of $\mathrm{A}$ is greater than the molar mass of B, then
ii. If the molar mass of $\mathrm{B}$ is greater than the molar mass of A, then
b. The products of the reaction are carbon dioxide (C) and water (D). Compound A has a similar molar mass to carbon dioxide. Compound B is a diatomic molecule. Identify compound $\mathrm{B},$ and support your answer.
c. Compound $A$ is a hydrocarbon that is $81.71 \%$ carbon by mass. Determine its empirical and molecular formulas.

Key Concepts

Molar Mass and Conversion

Molar mass is the mass of one mole of a substance and is used to convert between mass and moles. When equal masses of reactants with different molar masses are used, their mole quantities differ, which can affect the stoichiometric balance and determine which reactant is in excess or limiting.

Limiting Reactants

This concept involves determining which reactant in a chemical reaction will be completely consumed first, thereby limiting the amount of product that can be formed. It requires comparing the mole ratios of reactants, as derived from a balanced chemical equation, after converting the masses of reactants to moles, often influenced by their respective molar masses.

Stoichiometry

Stoichiometry is the calculation of reactants and products in chemical reactions based on a balanced chemical equation. It involves using the mole ratios derived from the equation to predict how much product will be formed from given quantities of reactants or to identify the limiting reactant.

Percent Composition and Empirical Formula Calculation

This concept entails determining the simplest whole-number ratio of atoms in a compound using the percentage composition by mass of each element. It involves converting percentages to masses, then to moles, and finally simplifying the mole ratios to obtain the empirical formula; additional steps may be needed to find the molecular formula if the molar mass is known.

Diatomic Molecules

Diatomic molecules consist of two atoms of the same element bonded together. Understanding diatomic molecules is important in recognizing naturally occurring elemental forms, and in this context, it helps in identifying a compound that exists as a diatomic species, affecting its molar mass and stoichiometry.

Transcript

00:01 But this problem, we're given a balanced equation of a plus 5b yields 3c plus 4d.

00:07 And we're told that equal masses, so the mass of a equals the mass of b, are reacted.

00:18 And we're trying to figure out which will be the limiting reactor based on two different scenarios.

00:23 In the first scenario, it says that if the molar mass of a is greater than the molar mass of b, can we determine what the limiting reactant is? and the answer is we cannot determine limiting reactant because even if a is greater than b, we cannot be sure that a will be greater than 5 times b.

01:19 So because we don't know that it will be greater than five moles of b, we cannot be sure if a will be a limiting reactant or if b will be the limiting reactant.

01:30 However, in the second scenario where it says that b, the molar mass of b is greater than the molar mass of a, then we can be sure that b is the limiting reactant because if if the molar mass of b is greater than a, then 5b is greater than a, so therefore that means that there will be fewer moles of b.

02:29 So it will be the limiting reactant in this scenario.

02:35 For the next part of the question, again, we have our equation of a plus 4.

02:41 5b produces 3c plus 4d, we are told that c is co2 and d is h2o.

02:56 So we can rewrite this equation as a plus 5b produces 3 co2 plus 4h2, which is a combustion reaction.

03:13 But they're asking us if a has a similar molar mass to carbon dioxide and b is a diatomic molecule, we want to identify b and support our answer.

03:24 Now, since it's the combustion reaction, the diatomic compound in combustion reactions is o2, but let's go ahead and prove that o2 is actually going to be what this is.

03:36 So first thing we want to do is find the molar mass of co2.

03:41 So co2 has a molar mass of 44 .01 grams per mole, and h2o has a molar mass of 18 .016 grams per mole.

03:59 So we can find that on our product side, we're going to have 3 times 44 .01 plus 4 times 18 .06 .6.

04:13 Which gives us a total mass of 204 .094.

04:23 And based upon the law of conservation of matter, if there are 204 .94 grams on the product side, that's how many have to be on the reactant side as well...

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