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Chapter 5: Problem 71

A \(15.0\) -L rigid container was charged with \(0.500\) atm of krypton gas and \(1.50\) atm of chlorine gas at \(350 .{ }^{\circ} \mathrm{C}\). The krypton and chlorine react to form krypton tetrachloride. What mass of krypton tetrachloride can be produced assuming \(100 \%\) yield?

Short Answer

Expert verified
The mass of krypton tetrachloride produced can be calculated by first finding the moles of krypton and chlorine using the ideal gas law, determining the limiting reactant, calculating the moles of krypton tetrachloride produced, and finally converting the moles to mass using the molar mass of KrCl4. The balanced chemical equation for this reaction is \( Kr + 4\ Cl_2 \rightarrow KrCl_4 \). The limiting reactant is determined by dividing the moles of each reactant by their respective coefficients in the balanced equation, and finding the smallest result. Assuming a 100% yield, the mass of krypton tetrachloride produced can be calculated as: \( Mass \, of \, KrCl_4 = moles \, of \, KrCl_4 \cdot molar \, mass \, of \, KrCl_4 \)

Step by step solution

01

Write the balanced chemical equation for the reaction

Krypton and chlorine react to form krypton tetrachloride, and the balanced chemical equation for the reaction is: \( Kr + 4\ Cl_2 \rightarrow KrCl_4 \)
02

Calculate the initial moles of reactants using the ideal gas law (PV=nRT)

The ideal gas law relates pressure, volume, temperature, and number of moles for a gas. Rearrange the ideal gas law to solve for n (moles): \( n = \frac{PV}{RT} \) Use the given pressure, volume, and temperature for each gas to calculate the initial moles. (Note: Temperature should be in Kelvin, so add 273.15 to the given temperature in Celsius) For krypton: \( n_{Kr} = \frac{(0.500 \ atm)(15.0 \ L)}{(0.0821 \ L \cdot atm \cdot mol^{-1} \cdot K^{-1})(350^{\circ}C + 273.15)} \) For chlorine: \( n_{Cl_2} = \frac{(1.50 \ atm)(15.0 \ L)}{(0.0821 \ L \cdot atm \cdot mol^{-1} \cdot K^{-1})(350^{\circ}C + 273.15)} \)
03

Determine the limiting reactant

To determine the limiting reactant, divide the moles of each reactant by their respective coefficients in the balanced chemical equation, and find the smallest result. For krypton: \( \frac{n_{Kr}}{1} \) For chlorine: \( \frac{n_{Cl_2}}{4} \) Whichever is smaller, that reactant is the limiting reactant.
04

Calculate the moles of krypton tetrachloride produced

Since we are assuming a 100% yield, the moles of krypton tetrachloride produced depends on the limiting reactant. Use the mole ratio from the balanced equation to calculate moles of krypton tetrachloride: \( moles \, KrCl_4 = moles \, limiting \, reactant \cdot \frac{1 \, mol \, KrCl_4}{1 \, mol \, limiting \, reactant} \)
05

Convert moles of krypton tetrachloride to mass

Now that we have the moles of krypton tetrachloride produced, we can find its mass using the molar mass of KrCl4: Molar mass of KrCl4 = M(Kr) + 4 * M(Cl) = 83.80 g/mol + 4 * 35.45 g/mol Mass of KrCl4 = moles of KrCl4 * molar mass of KrCl4

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Most popular questions from this chapter

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