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

For each of the following unbalanced chemical equations, suppose that exactly 5.00 g of each reactant is taken. Determine which reactant is limiting, and calculate what mass of each product is expected (assuming that the limiting reactant is completely consumed). a. $\mathrm{S}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \rightarrow \mathrm{SO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l)$ b. $\mathrm{MnO}_{2}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(l) \rightarrow \operatorname{Mn}\left(\mathrm{SO}_{4}\right)_{2}(s)+\mathrm{H}_{2} \mathrm{O}(l)$ c. $\mathrm{H}_{2} \mathrm{S}(g)+\mathrm{O}_{2}(g) \rightarrow \mathrm{SO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l)$ d. $\mathrm{AgNO}_{3}(a q)+\mathrm{Al}(s) \rightarrow \mathrm{Ag}(s)+\mathrm{Al}\left(\mathrm{NO}_{3}\right)_{3}(a q)$

For each of the following unbalanced chemical equations, suppose that exactly 5.00 g of each reactant is taken. Determine which reactant is limiting, and calculate what mass of each product is expected (assuming that the limiting reactant is completely consumed).
a. $\mathrm{S}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \rightarrow \mathrm{SO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l)$
b. $\mathrm{MnO}_{2}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(l) \rightarrow \operatorname{Mn}\left(\mathrm{SO}_{4}\right)_{2}(s)+\mathrm{H}_{2} \mathrm{O}(l)$
c. $\mathrm{H}_{2} \mathrm{S}(g)+\mathrm{O}_{2}(g) \rightarrow \mathrm{SO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l)$
d. $\mathrm{AgNO}_{3}(a q)+\mathrm{Al}(s) \rightarrow \mathrm{Ag}(s)+\mathrm{Al}\left(\mathrm{NO}_{3}\right)_{3}(a q)$

Key Concepts

Balanced Chemical Equations

Balanced chemical equations are essential for understanding and applying the law of conservation of mass in chemical reactions. They provide the mole ratios in which reactants combine and products form, ensuring that the number of atoms for each element is the same on both sides of the reaction. This balance allows chemists to accurately calculate the quantities of reactants needed and the amounts of products produced.

Limiting Reactant

The limiting reactant is the substance in a chemical reaction that is completely consumed first, thus determining the maximum amount of product that can be formed. Identifying the limiting reactant is critical in stoichiometry because it limits the extent of the reaction and the yield of the desired products.

Conversion Between Mass and Moles

Converting between mass and moles is a fundamental concept in chemistry used to relate the physical amount of a substance to the number of particles it contains. This conversion is carried out using the molar mass of the substance, enabling one to use balanced chemical equations to determine how much of a reactant is required or how much of a product will be formed.

Stoichiometry Calculations

Stoichiometry involves using the relationships derived from balanced chemical equations to compute the quantities of reactants and products involved in a chemical reaction. It incorporates mole ratios, mass-mole conversions, and the concept of the limiting reactant to predict yields and assess the efficiency of chemical processes.

Transcript

00:01 That got full so i'm going to, fortunately i've done this, i'm going to solve this with these.

00:12 Okay, so we're given the following information.

00:17 Suppose 5 grams of each reactant is provided.

00:21 We are asked to determine the limiting reactant and the mass of each product formed.

00:28 Here's my first reaction and i've balanced it.

00:32 It was unbalanced.

00:34 We're given this reaction.

00:36 So i've set up my two calculations, each starting with 5 moles of my given.

00:41 I use the molar mass of sulfuric acid and then i multiply that by the mole ratio of 3 moles of so2 to 2 moles of sulfuric acid.

00:53 Finally, i multiply by the molar mass of sulfur dioxide.

00:57 My answer is 4 .9 grams of sulfur dioxide.

01:02 My second calculation is 4 or 5 grams of sulfur times 32 .06 grams of sulfur.

01:12 Then we're going to multiply it by the mole ratio of 3 to 1.

01:17 And finally, again, every once in a while when i do these, by the molar mass.

01:25 I see that i've made some silly little mistake or omission.

01:29 This should be moles of sulfur.

01:36 There we go.

01:37 And this is 30 .00 grams of so2.

01:41 I just quit writing.

01:42 There we go.

01:47 And i got 30 grams.

01:49 From that information, we can see that my limiting reactant is sulfuric acid and i have 4 .90 grams of sulfuric acid produced.

01:58 Now, we can take our limiting reactant, again multiply by its molar mass.

02:04 Then the mole ratio for water to sulfuric acid and the molar mass of water.

02:10 And now i have the answers for all three of these.

02:15 I've got, let's see, i was going to go into a pen for this.

02:21 Here's my limiting reactant and sulfur dioxide produced and the amount of water produced.

02:31 Next, let's take a look at our next problem.

02:39 Here, whoops, here is our balanced chemical equation.

02:51 Now we're going to take 5 grams of each reactant.

02:54 This time i did it a little bit differently.

02:56 I converted each to moles.

02:58 There's moles of each reactant.

03:01 Then i took moles of the mno4 and i multiplied this mno2...

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