A sample of nitrosyl bromide (NOBr) decomposes according to the equation 2 NOBr(g) ⇌ 2 NO(g) + Br₂(g) An equilibrium mixture in a 5.00-L vessel at 100°C contains 3.22 g of NOBr, 3.08 g of NO, and 4.19 g of Br₂. (c) What was the mass of the original sample of NOBr?

검증된 단계별 안내

Calculate the molar mass of NOBr using the atomic masses of nitrogen (N), oxygen (O), and bromine (Br).

Convert the given mass of NOBr at equilibrium (3.22 g) to moles using its molar mass.

Convert the given masses of NO (3.08 g) and Br₂ (4.19 g) to moles using their respective molar masses.

Use the stoichiometry of the balanced equation 2 NOBr ⇌ 2 NO + Br₂ to determine the moles of NOBr that decomposed to form the moles of NO and Br₂ at equilibrium.

Add the moles of NOBr at equilibrium to the moles of NOBr that decomposed to find the initial moles of NOBr, then convert this back to grams to find the original mass of the NOBr sample.

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주요 개념

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Stoichiometry

Stoichiometry is the calculation of reactants and products in chemical reactions based on the balanced chemical equation. In this case, the decomposition of nitrosyl bromide (NOBr) involves a specific molar ratio of reactants to products, which is essential for determining the amounts of substances involved in the reaction.

Equilibrium

Chemical equilibrium occurs when the rates of the forward and reverse reactions are equal, resulting in constant concentrations of reactants and products. Understanding equilibrium is crucial for analyzing the decomposition of NOBr, as it helps in determining how much of the original sample remains and how much has reacted at a given time.

Molar Mass

Molar mass is the mass of one mole of a substance, typically expressed in grams per mole. To find the mass of the original sample of NOBr, it is necessary to calculate the number of moles present in the equilibrium mixture using the molar mass of NOBr, which allows for the determination of the initial mass before decomposition.

A sample of nitrosyl bromide (NOBr) decomposes according to the equation 2 NOBr(g) ⇌ 2 NO(g) + Br2(g) An equilibrium mixture in a 5.00-L vessel at 100°C contains 3.22 g of NOBr, 3.08 g of NO, and 4.19 g of Br2. (c) What was the mass of the original sample of NOBr?