Challenge Balance the following equations, and determine the...

Challenge Balance the following equations, and determine the possible mole ratios. \begin{equation} \begin{array}{l}{\text { a. } \mathrm{ZnO}(\mathrm{s})+\mathrm{HCl}(\mathrm{aq}) \rightarrow \mathrm{ZnCl}_{2}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l})} \\ {\text { b. butane }\left(\mathrm{C}_{4} \mathrm{H}_{10}\right)+\text { oxygen } \rightarrow \text { carbon dioxide }+\text { water }}\end{array} \end{equation}

Challenge Balance the following equations, and determine the possible mole ratios.
\begin{equation}
\begin{array}{l}{\text { a. } \mathrm{ZnO}(\mathrm{s})+\mathrm{HCl}(\mathrm{aq}) \rightarrow \mathrm{ZnCl}_{2}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l})} \\ {\text { b. butane }\left(\mathrm{C}_{4} \mathrm{H}_{10}\right)+\text { oxygen } \rightarrow \text { carbon dioxide }+\text { water }}\end{array}
\end{equation}

Key Concepts

Conservation of Mass

The principle of conservation of mass states that matter is neither created nor destroyed in a chemical reaction. This concept underpins the need for balanced chemical equations because it requires that the total mass of the reactants equals the total mass of the products.

Balancing Chemical Equations

This key concept involves adjusting the coefficients of reactants and products to ensure that the number of atoms for each element is the same on both sides of the equation. It is essential in chemistry to reflect the law of conservation of mass and accurately represent the reaction taking place.

Mole Ratios

Mole ratios are derived from the coefficients in a balanced chemical equation and indicate the relative number of moles of each substance involved in the reaction. They provide the basis for converting between amounts of reactants and products, which is crucial for quantitative chemical calculations.

Stoichiometry

Stoichiometry is the branch of chemistry that deals with the calculations relating to the quantities of reactants and products in a chemical reaction. It uses balanced chemical equations to determine the amounts of substances consumed and produced, ensuring that the proportions of each are maintained according to their mole ratios.

Transcript

00:01 Balance these two equations and then determine some possible molar ratios.

00:04 Let's start with equation a, where we have zinc oxide reacting with hydrochloric acid, forming zinc chloride and water.

00:12 Let's add our states and our zinc oxide is going to be a solid.

00:22 So let's take a look at our zinc.

00:24 Notice we have one on the left, one on the right.

00:26 That is good to go.

00:27 Oxygen, one on the left, one on the right.

00:29 Good to go.

00:31 Our hydrogen and our chlorine, however, they're two on the right, one on the left.

00:35 So we just have to add a two in front of the hydrochloric acid.

00:39 So if you're going to be comparing zinc oxide to hydrochloric acid, the ratio is going to be a one to two.

00:47 And if you're comparing zinc oxide to zinc chloride, the reaction is going to be a one to one molar ratio and hydrochloric acid to zinc chloride.

01:00 Well, that's going to be a two to one.

01:03 And finally, if you wanted to compare hydrochloric acid to water, i'm going to rewrite hydrochloric acid because that just looks weird.

01:16 Well, that ratio is going to be a two to one.

01:19 You can go even further.

01:20 You can go into ions, but i'm going to stop here with just the, or with the compounds.

01:25 Now let's take a look at equation b, where we have butane reacting with excess oxygen to form carbon dioxide and water.

01:37 And i went ahead and i converted everything into the formulas.

01:41 So butane, c4h10, oxygen is o2.

01:45 Remember oxygen is diatomic, carbon dioxide, co2, water, h2o.

01:50 So i'm going to set up a little table where i have c, h, and o listed on both sides.

01:57 I'm going to write down our starting numbers of moles.

02:01 We have four of carbon, 10 of oxygen, or sorry, 10 of hydrogen, two of oxygen on the left.

02:07 And on the right, we have one carbon, we have two hydrogen, and we have two plus one oxygen.

02:13 It's tempting to combine those into a three, but keeping them separate allows you to balance them separately.

02:19 So we're going to do that.

02:22 Notice our carbon, we have four on the left, one on the right.

02:24 So immediately we're going to have to multiply that carbon by four.

02:28 That means we're going to multiply that oxygen by four.

02:31 So now this becomes eight plus one.

02:35 Next, we're going to go to our hydrogens...

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