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Chapter 9: Problem 138

Potassium nitrate decomposes upon heating to form potassium oxide, nitrogen gas, and oxygen gas. If \(19.6 \mathrm{~g}\) of potassium nitrate have decomposed, how many molecules of oxygen gas have been formed?

Short Answer

Expert verified
The number of oxygen gas molecules formed in the decomposition of 19.6 g of potassium nitrate is approximately \(2.92 \times 10^{23}\) molecules.

Step by step solution

01

Write the balanced chemical equation for the decomposition of potassium nitrate

The decomposition reaction of potassium nitrate is given as follows: 2 KNO₃ (s) → 2 K₂O (s) + 2 N₂ (g) + 5 O₂ (g)
02

Calculate the moles of potassium nitrate (KNO₃)

Given, 19.6 g of potassium nitrate decomposes. First, calculate the molar mass of KNO₃ using the atomic masses of its constituent elements: Molar mass of KNO₃ = (1 × atomic mass of K) + (1 × atomic mass of N) + (3 × atomic mass of O) Molar mass of KNO₃ = (1 × 39.1 g/mol) + (1 × 14.0 g/mol) + (3 × 16.0 g/mol) Molar mass of KNO₃ = 39.1 + 14.0 + 48.0 Molar mass of KNO₃ = 101.1 g/mol Now, calculate the moles of KNO₃ using its given mass and molar mass: Moles of KNO₃ = Mass of KNO₃ / Molar mass of KNO₃ Moles of KNO₃ = 19.6 g / 101.1 g/mol Moles of KNO₃ = 0.194 mol
03

Use stoichiometry to find moles of oxygen gas (O₂) formed

According to the balanced chemical equation, 2 moles of KNO₃ produce 5 moles of O₂. Therefore, we can set up the following proportion to find the moles of O₂ produced: (0.194 mol KNO₃)(5 mol O₂ / 2 mol KNO₃) Now, calculate the moles of O₂: Moles of O₂ = 0.194 × (5/2) Moles of O₂ = 0.485 mol
04

Calculate the number of molecules of oxygen gas from moles

The number of molecules can be calculated using Avogadro's constant: Number of molecules = Moles of O₂ × (6.022 × 10²³ molecules/mol) Number of molecules = 0.485 mol × (6.022 × 10²³) Number of molecules = 2.92 × 10²³ molecules The number of oxygen gas molecules formed is approximately \(2.92 \times 10^{23}\) molecules.

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Most popular questions from this chapter

Tetraphosphorus decoxide, \(\mathrm{P}_{4} \mathrm{O}_{10}\), reacts with water to form phosphoric acid. If \(52.5 \mathrm{~g}\) of \(\mathrm{P}_{4} \mathrm{O}_{10}\) reacted with \(25.0 \mathrm{~g}\) of water, how many grams of \(\mathrm{H}_{3} \mathrm{PO}_{4}\) could theoretically be produced?

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Ethylene glycol, used for antifreeze, has the molecular formula \(\mathrm{C}_{2} \mathrm{H}_{6} \mathrm{O}_{2}\). Calculate the mass percent of each element in ethylene glycol.

Consider the balanced chemical equation \(2 \mathrm{~A}+\mathrm{B} \rightarrow 2 \mathrm{C}+\mathrm{D}\) When \(8.0 \mathrm{~g}\) of A reacts completely with \(6.0 \mathrm{~g}\) of \(\mathrm{B}\), \(10.0 \mathrm{~g}\) of \(\mathrm{C}\) and \(4.0 \mathrm{~g}\) of \(\mathrm{D}\) are produced. Assuming the yield is \(100 \%\), (a) Which has a greater molar mass, A or C? (b) Which has a greater molar mass, A or B? (c) Which has a greater molar mass, A or D? (d) If the molar mass of \(\mathrm{A}\) is \(24.0 \mathrm{~g} / \mathrm{mol}\), determine the molar mass of \(\mathrm{B}, \mathrm{C}\), and \(\mathrm{D}\).

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