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Chapter 5: Problem 124

At STP, \(1.0 \mathrm{~L} \mathrm{Br}_{2}\) reacts completely with \(3.0 \mathrm{~L} \mathrm{~F}_{2}\), producing 2.0 L of a product. What is the formula of the product? (All substances are gases.)

Short Answer

Expert verified
The formula of the product is \( BrF_{3} \).

Step by step solution

01

1. Write down the unbalanced chemical equation

Based on the information given, we can write down the unbalanced chemical equation: \[ Br_{2} + F_{2} \rightarrow ? \]
02

2. Write the balanced chemical equation

We know that the reaction occurs at STP, which means that both reactants and products behave as ideal gases. The balanced chemical equation for the given reaction can be derived as follows: \[ Br_{2} + 3F_{2} \rightarrow 2BrF_{3} \] This equation indicates that one mole of bromine reacts with three moles of fluorine to produce two moles of the product.
03

3. Calculate the mole ratio of reactants:

At STP, the volume of a gas is directly proportional to the number of moles. Therefore, the mole ratio of the reactants is: \( \frac{n_{Br_{2}}}{n_{F_{2}}} = \frac{1.0 L}{3.0 L} = \frac{1}{3} \) where \( n_{Br_{2}} \) and \( n_{F_{2}} \) represent the moles of bromine and fluorine, respectively.
04

4. Compare the calculated mole ratio with the balanced chemical equation

According to the balanced chemical equation, 1 mole of bromine reacts with 3 moles of fluorine: \( \frac{n_{Br_{2}}}{n_{F_{2}}} = \frac{1}{3} \) This is the same as the calculated mole ratio of the reactants.
05

5. Determine the formula of the product

As we have the stoichiometry of the reaction from the balanced chemical equation and the volume of product formed, we can use that to determine the formula of the product. The balanced chemical equation: \[ Br_{2} + 3F_{2} \rightarrow 2BrF_{3} \] Since \(2.0 L\) of a product is formed, and we have this relationship from the balanced equation: \( 2 \: moles \: Product = 1.0 \: L \: Br_{2} \: reactant \) Thus, the product formed is \( BrF_{3} \). The formula of the product is \( BrF_{3} \).

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Most popular questions from this chapter

Methanol \(\left(\mathrm{CH}_{3} \mathrm{OH}\right)\) can be produced by the following reaction: $$ \mathrm{CO}(g)+2 \mathrm{H}_{2}(g) \longrightarrow \mathrm{CH}_{3} \mathrm{OH}(g) $$ Hydrogen at STP flows into a reactor at a rate of \(16.0 \mathrm{~L} / \mathrm{min}\) Carbon monoxide at STP flows into the reactor at a rate of \(25.0\) L/min. If \(5.30 \mathrm{~g}\) methanol is produced per minute, what is the percent yield of the reaction?

Hydrogen azide, \(\mathrm{HN}_{3}\), decomposes on heating by the following unbalanced reaction: $$ \mathrm{HN}_{3}(g) \longrightarrow \mathrm{N}_{2}(g)+\mathrm{H}_{2}(g) $$ If \(3.0 \mathrm{~atm}\) of pure \(\mathrm{HN}_{3}(\mathrm{~g})\) is decomposed initially, what is the final total pressure in the reaction container? What are the partial pressures of nitrogen and hydrogen gas? Assume the volume and temperature of the reaction container are constant.

Equal moles of sulfur dioxide gas and oxygen gas are mixed in a flexible reaction vessel and then sparked to initiate the formation of gaseous sulfur trioxide. Assuming that the reaction goes to completion, what is the ratio of the final volume of the gas mixture to the initial volume of the gas mixture if both volumes are measured at the same temperature and pressure?

Ethene is converted to ethane by the reaction $$ \mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{H}_{2}(g) \stackrel{\mathrm{Cindys}}{\longrightarrow} \mathrm{C}_{2} \mathrm{H}_{6}(g) $$ \(\mathrm{C}_{2} \mathrm{H}_{4}\) flows into a catalytic reactor at \(25.0 \mathrm{~atm}\) and \(300 .{ }^{\circ} \mathrm{C}\) with a flow rate of \(1000 .\) L/min. Hydrogen at \(25.0\) atm and \(300 .{ }^{\circ} \mathrm{C}\) flows into the reactor at a flow rate of \(1500 . \mathrm{L} / \mathrm{min}\). If \(15.0 \mathrm{~kg} \mathrm{C}_{2} \mathrm{H}_{6}\) is collected per minute, what is the percent yield of the reaction?

Metallic molybdenum can be produced from the mineral molybdenite, \(\mathrm{MoS}_{2}\). The mineral is first oxidized in air to molybdenum trioxide and sulfur dioxide. Molybdenum trioxide is then reduced to metallic molybdenum using hydrogen gas. The balanced equations are $$ \begin{aligned} \mathrm{MoS}_{2}(s)+\frac{2}{2} \mathrm{O}_{2}(g) & \longrightarrow \mathrm{MoO}_{3}(s)+2 \mathrm{SO}_{2}(g) \\ \mathrm{MoO}_{3}(s)+3 \mathrm{H}_{2}(g) & \longrightarrow \mathrm{Mo}(s)+3 \mathrm{H}_{2} \mathrm{O}(l) \end{aligned} $$ Calculate the volumes of air and hydrogen gas at \(17^{\circ} \mathrm{C}\) and \(1.00\) atm that are necessary to produce \(1.00 \times 10^{3} \mathrm{~kg}\) pure molybdenum from \(\mathrm{MoS}_{2}\). Assume air contains \(21 \%\) oxygen by volume and assume \(100 \%\) yield for each reaction.
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