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Chapter 3: Problem 100

The reaction between potassium chlorate and red phosphorus takes place when you strike a match on a matchbox. If you were to react \(52.9 \mathrm{~g}\) of potassium chlorate \(\left(\mathrm{KClO}_{3}\right)\) with excess red phosphorus, what mass of tetraphosphorus decaoxide \(\left(\mathrm{P}_{4} \mathrm{O}_{10}\right)\) would be produced? $$ \mathrm{KClO}_{3}(s)+\mathrm{P}_{4}(s) \longrightarrow \mathrm{P}_{4} \mathrm{O}_{10}(s)+\mathrm{KCl}(s) \quad \text { (unbalanced) } $$

Short Answer

Expert verified
When 52.9 g of potassium chlorate reacts with excess red phosphorus, 20.4 g of tetraphosphorus decaoxide (P4O10) is produced.

Step by step solution

01

Balance the chemical equation

To balance the chemical equation, we need to make sure there are equal numbers of atoms of each element on both sides of the equation. The balanced equation is: $$ 6 \mathrm{KClO}_{3}(s)+\mathrm{P}_{4}(s) \longrightarrow \mathrm{P}_{4}\mathrm{O}_{10}(s)+6\mathrm{KCl}(s) $$
02

Finding moles of KClO3

First we need to find the moles of potassium chlorate (KClO3) given that we have 52.9 g of it. The molar mass of KClO3 is calculated as follows: Molar mass (KClO3) = 39.10(g/mol for K) + 35.45(g/mol for Cl) + 3 * 16(g/mol for O) = 122.55 g/mol Now, we can find the moles of KClO3: Moles(KClO3) = \(\frac{52.9 \mathrm{~g}}{122.55 \mathrm{~g/mol}}\) = 0.432 moles
03

Using stoichiometry to find moles of P4O10 produced

Now, we will use stoichiometry to find the moles of tetraphosphorus decaoxide (P4O10) produced. According to the balanced chemical equation, 6 moles of KClO3 react with 1 mole of P4O10. We can write this ratio as: Moles(P4O10) = Moles(KClO3) * \(\frac{1 \text{ mole } \mathrm{P}_{4}\mathrm{O}_{10}}{6 \text{ moles } \mathrm{KClO}_{3}}\) Moles(P4O10) = 0.432 moles * \(\frac{1}{6}\) = 0.072 mol
04

Calculate mass of P4O10 produced

Now, we will calculate the mass of P4O10 produced using its molar mass. The molar mass of P4O10 is calculated as follows: Molar mass (P4O10) = 4 * 30.97(g/mol for P) + 10 * 16(g/mol for O) = 283.88 g/mol To find the mass of P4O10 produced: Mass(P4O10) = Moles(P4O10) * Molar mass(P4O10) Mass(P4O10) = 0.072 mol * 283.88 g/mol = 20.4 g So, when 52.9 g of potassium chlorate reacts with excess red phosphorus, 20.4 g of tetraphosphorus decaoxide (P4O10) is produced.

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Most popular questions from this chapter

The compound adrenaline contains \(56.79 \%\) C, \(6.56 \% \mathrm{H}\). \(28.37 \%\) O, and \(8.28 \%\) N by mass. What is the empirical formula for adrenaline?

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Consider the following data for three binary compounds of hydrogen and nitrogen: $$ \begin{array}{lcc} & \% \mathrm{H} \text { (by Mass) } & \text { \% N (by Mass) } \\ \hline \text { I } & 17.75 & 82.25 \\ \text { II } & 12.58 & 87.42 \\ \text { III } & 2.34 & 97.66 \end{array} $$ When \(1.00 \mathrm{~L}\) of each gaseous compound is decomposed to its elements, the following volumes of \(\mathrm{H}_{2}(g)\) and \(\mathrm{N}_{2}(g)\) are obtained: $$ \begin{array}{lcc} & \mathrm{H}_{2} \text { (L) } & \mathrm{N}_{2} \text { (L) } \\ \hline \text { I } & 1.50 & 0.50 \\ \text { II } & 2.00 & 1.00 \\ \text { III } & 0.50 & 1.50 \end{array} $$ Use these data to determine the molecular formulas of compounds I, II, and III and to determine the relative values for the atomic masses of hydrogen and nitrogen.

A \(0.755-\mathrm{g}\) sample of hydrated copper(II) sulfate $$ \mathrm{CuSO}_{4} \cdot x \mathrm{H}_{2} \mathrm{O} $$ was heated carefully until it had changed completely to anhydrous copper(II) sulfate \(\left(\mathrm{CuSO}_{4}\right)\) with a mass of \(0.483 \mathrm{~g}\). Determine the value of \(x\). [This number is called the number of waters of hydration of copper(II) sulfate. It specifies the number of water molecules per formula unit of \(\mathrm{CuSO}_{4}\) in the hydrated crystal.]
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