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Problem 68

Chlorine gas can be made in the laboratory by the reaction of hydrochloric acid and manganese(IV) oxide:

$$

4 \mathrm{HCl}(a q)+\mathrm{MnO}_{2}(s) \longrightarrow \mathrm{MnCl}_{2}(a q)+2 \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{Cl}_{2}(g)

$$

When 1.82 mol of $\mathrm{HCl}$ reacts with excess $\mathrm{MnO}_{2},$ how many (a) moles of $\mathrm{Cl}_{2}$ and (b) grams of $\mathrm{Cl}_{2}$ form?

Answer

$\mathrm{n}\left(\mathrm{Cl}_{2}\right)=0,455 \mathrm{mol}$

$\mathrm{m}\left(\mathrm{Cl}_{2}\right)=32,26 \mathrm{g}$

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## Discussion

## Video Transcript

we have the reaction between the electric or acid and and then Oh, for Michael. So we have one point you to move off its CEO being going toe, We ever accept our manganese oxide? Damn reporters are Megan's core I water and Corey. So ah, for party. We're going to find out how What will be the limbo, More Caribbean, 40 years. And also what with the mass off Cohen being for deals. Okay, so for these questions, first of all, we have to make sure we have violence. And, um actually, we have to cancel with immigration bothers. So from there, from the coefficient of the chemical reaction ah, we can find out their corresponding motorway show. Well, the way show. Okay. So Ah, little at corps. Um hi, Joe. Cory Acid. We're four, and then we want to find out. Ah, Korean gas. The number more chlorine gas. So we look at that. It's one. So therefore, from here, the mullahs wish off hydrochloride acid to Korean gas will be close to force her 4 to 1. All right, so if the lumber most off a hydrochloric acid eyes one point a to mo. So what would be the lumber. Almost all Korean beanpole. Adios. So we know that the more you ratio is 4 to 1. So we just divide that by four, and then we should be able to find a sitter for 455 moles for Cohen and gas being poor deals. Okay, so this will be for A and M bonfire. Be the mass of Corinne gas. It will be. Ah, we're looking at Cohen Gas, and we just take the Lambo most with applied by the well, the mass off calling Cohen is a gas eyes. See out you. So come on. By that. Ah! 35.4 500 of the mass. Times two. And they were We have the masses. Your soup on 455 Time the off. Ah, 35.45 times two. So we would have furry two point, um, two crab for the myself. Korean being buddies

## Recommended Questions

Chlorine forms from the reaction of hydrochloric acid with manganese(IV) oxide. The balanced equation is:

\begin{equation}

\mathrm{MnO}_{2}+4 \mathrm{HCl} \rightarrow \mathrm{MnCl}_{2}+\mathrm{Cl}_{2}+2 \mathrm{H}_{2} \mathrm{O}

\end{equation}

Calculate the theoretical yield and the percent yield of

chlorine if 86.0 $\mathrm{g}$ of $\mathrm{MnO}_{2}$ and 50.0 $\mathrm{g}$ of $\mathrm{HCl}$ react. The

actual yield of $\mathrm{Cl}_{2}$ is 20.0 $\mathrm{g}$ .

Consider the reaction

$$\mathrm{MnO}_{2}+4 \mathrm{HCl} \longrightarrow \mathrm{MnCl}_{2}+\mathrm{Cl}_{2}+2 \mathrm{H}_{2} \mathrm{O}$$

If 0.86 mole of $\mathrm{MnO}_{2}$ and $48.2 \mathrm{g}$ of $\mathrm{HCl}$ react, which reactant will be used up first? How many grams of $\mathrm{Cl}_{2}$ will be produced?

Consider the reaction

$$

\mathrm{MnO}_{2}+4 \mathrm{HCl} \longrightarrow \mathrm{MnCl}_{2}+\mathrm{Cl}_{2}+2 \mathrm{H}_{2} \mathrm{O}

$$

If 0.86 mole of $\mathrm{MnO}_{2}$ and $48.2 \mathrm{g}$ of $\mathrm{HCl}$ react, which reagent will be used up first? How many grams of $\mathrm{Cl}_{2}$ will be produced?

Chlorine can be generated by heating together calcium hypochlorite and hydrochloric acid. Calcium chloride and water are also formed. (a) If $50.0 \mathrm{g}$ $\mathrm{Ca}(\mathrm{OCl})_{2}$ and $275 \mathrm{mL}$ of $6.00 \mathrm{M} \mathrm{HCl}$ are allowed to

react, how many grams of chlorine gas will form? (b) Which reactant, $\mathrm{Ca}(\mathrm{OCl})_{2}$ or $\mathrm{HCl}$, remains in excess, and in what mass?

Hydrogen and chlorine react to yield hydrogen chloride: $\mathrm{H}_{2}+\mathrm{Cl}_{2} \rightarrow 2 \mathrm{HCl}$ . How many grams of $\mathrm{HCl}$ are formed from reaction of 3.56 $\mathrm{g}$ of $\mathrm{H}_{2}$ with 8.94 $\mathrm{g}$ of $\mathrm{Cl}_{2} ?$ Which reactant is limiting?

Consider the reaction below.

$\mathrm{Ca}(\mathrm{OH})_{2}(\mathrm{s})+2 \mathrm{HCl}(\mathrm{aq}) \longrightarrow \mathrm{CaCl}_{2}(\mathrm{aq})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})$

(a) How many grams of $\mathrm{Ca}(\mathrm{OH})_{2}$ are required to react completely with $415 \mathrm{mL}$ of $0.477 \mathrm{M} \mathrm{HCl} ?$

(b) How many kilograms of $\mathrm{Ca}(\mathrm{OH})_{2}$ are required to react with 324 L of a HCl solution that is $24.28 \%$ HCl by mass, and has a density of $1.12 \mathrm{g} / \mathrm{mL}$ ?

How many moles of HCl can be made from 6.15 $\mathrm{mol}$ $\mathrm{H}_{2}$ and an excess of $\mathrm{Cl}_{2} ?$

The reaction below can be used as a laboratory method of preparing small quantities of $\mathrm{Cl}_{2}(\mathrm{g}) .$ If a 62.6 g sample that is 98.5\% K_CraO_ by mass is allowed to react with $325 \mathrm{mL}$ of $\mathrm{HCl}(\mathrm{aq})$ with a density of $1.15 \mathrm{g} / \mathrm{mL}$ and $30.1 \% \mathrm{HCl}$ by mass, how many grams of $\mathrm{Cl}_{2}(\mathrm{g})$ are produced?

$$

\begin{aligned}

\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+\mathrm{H}^{+}+\mathrm{Cl}^{-} & \longrightarrow \\

\mathrm{Cr}^{3+}+\mathrm{H}_{2} \mathrm{O} &+\mathrm{Cl}_{2}(\mathrm{g}) & \text { (not balanced) }

\end{aligned}

$$

Consider the reaction:

$$4 \mathrm{HCl}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(g)+2 \mathrm{Cl}_{2}(g)$$

Each molecular diagresents an initial mixture of the reactants.

How many molecules of $\mathrm{Cl}_{2}$ would from from the reaction mixture that

produces the greatest amount of products?

Given a quantity in moles of reactant or product, use a mole-mole factor from the balanced chemical equation to calculate the number of moles of another substance in the reaction.

Write all the mole-mole factors for each of the following chemical equations:

a. $2 \mathrm{Al}(s)+3 \mathrm{Cl}_{2}(g) \longrightarrow 2 \mathrm{AlCl}_{3}(s)$

b. $4 \mathrm{HCl}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{Cl}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g)$