For each of the following reactions, calculate the grams of...

For each of the following reactions, calculate the grams of indicated product when $25.0 \mathrm{~g}$ of the first reactant and $40.0 \mathrm{~g}$ of the second reactant are used: a. $2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{SO}_{3}(g) \quad\left(\mathrm{SO}_{3}\right)$ b. $3 \mathrm{Fe}(s)+4 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Fe}_{3} \mathrm{O}_{4}(s)+4 \mathrm{H}_{2}(g) \quad\left(\mathrm{Fe}_{3} \mathrm{O}_{4}\right)$ c. $\mathrm{C}_{7} \mathrm{H}_{16}(l)+11 \mathrm{O}_{2}(g) \stackrel{\Delta}{\longrightarrow} 7 \mathrm{CO}_{2}(g)+8 \mathrm{H}_{2} \mathrm{O}(g) \quad\left(\mathrm{CO}_{2}\right)$

For each of the following reactions, calculate the grams of indicated product when $25.0 \mathrm{~g}$ of the first reactant and $40.0 \mathrm{~g}$ of the second reactant are used:
a. $2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{SO}_{3}(g) \quad\left(\mathrm{SO}_{3}\right)$
b. $3 \mathrm{Fe}(s)+4 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Fe}_{3} \mathrm{O}_{4}(s)+4 \mathrm{H}_{2}(g) \quad\left(\mathrm{Fe}_{3} \mathrm{O}_{4}\right)$
c. $\mathrm{C}_{7} \mathrm{H}_{16}(l)+11 \mathrm{O}_{2}(g) \stackrel{\Delta}{\longrightarrow} 7 \mathrm{CO}_{2}(g)+8 \mathrm{H}_{2} \mathrm{O}(g) \quad\left(\mathrm{CO}_{2}\right)$

Key Concepts

Balanced Chemical Equations

Balanced chemical equations express the conservation of mass by showing the precise mole ratios in which reactants combine and products form. They provide the necessary proportional relationships used to relate the quantities of substances involved in a reaction.

Mole Concept

The mole is a fundamental unit in chemistry that quantifies a specific number of particles (Avogadro's number) and serves as a bridge between the microscopic world and measurable quantities. Using moles allows chemists to accurately relate masses to the number of atoms or molecules.

Molar Mass

Molar mass is the mass of one mole of a substance, typically expressed in grams per mole. This concept is crucial for converting between the mass of a substance and the number of moles, facilitating accurate stoichiometric calculations.

Stoichiometric Calculations

Stoichiometric calculations involve determining the quantitative relationships between reactants and products in a chemical reaction. By using the mole ratios from the balanced equation, it is possible to compute how much of a reactant is needed or how much product will be formed.

Limiting Reactant

The limiting reactant is the substance in a chemical reaction that is entirely consumed first, limiting the extent of the reaction and determining the maximum amount of product that can be produced. Identifying the limiting reactant is key to predicting yields in chemical reactions.

Transcript

00:01 For each of the following reactions, calculate the grams of the indicated product when 25 grams of the first reactant and 40 grams of the second reactant are used.

00:09 So this is a limiting reactant problem and in order to do these you needed to know or you needed to calculate the molecular weights.

00:17 I've already done that for each one of the reactants and the products.

00:21 So for part a, we're going to start with so2 and we're going to say we have 25 grams of that.

00:28 We're going to convert it to moles by saying one mole over the molecular weight which is 64 .06 grams.

00:38 From here we convert it from moles of reactant to moles of product by saying two moles of so2 will yield two moles of so3 and then from here you go one mole of so3 will yield or weighs 80 .06 grams and you do the same thing for the oxygen except this time you have 40 .0 grams times one mole of o2 weighs 32 grams.

01:15 Here the ratio of oxygen on the bottom is one mole of oxygen to two moles of so3 and then 80 .06 grams times one mole of so3.

01:32 The limiting reactant is going to be the reactant that produces the fewest number of moles and thus the fewest number of grams of product.

01:41 So we're going to start with so2 and we're going to write 25 and then divide that by 64 .06, multiply that by 2 over 2 and then multiply that by 80 .06 and this is going to yield 31 .24 grams of so3.

02:00 On the bottom we have 40 divided by 32 times 2 times 80 .06.

02:08 Here this is going to be 200 .15 grams.

02:13 So the limiting reactant in this case is going to be the so2 and it produces 31 .24 grams of the product.

02:27 Now let's take a look at b.

02:29 Here we have iron at 25 grams that's going to be multiplied by one mole over 55 .845 grams.

02:40 From here the ratio is three moles of iron to one mole of our product fe3o4 and from that one mole of that weighs 231 .535 grams.

03:00 Now for the h2o, the water, that has 40 .0 grams times one mole over 18 .016.

03:13 Here the ratio is four moles of water on the bottom to one mole of product, that should be a three on the top, and this is 231 .535 grams over one mole.

03:31 So let's start with the iron...

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