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Chapter 5: Problem 134

A 2.747 -g sample of manganese metal is reacted with excess \(\mathrm{HCl}\) gas to produce 3.22 \(\mathrm{L} \mathrm{H}_{2}(g)\) at 373 \(\mathrm{K}\) and 0.951 atm and a manganese chloride compound $\left(\mathrm{Mn} \mathrm{Cl}_{x}\right)$. What is the formula of the manganese chloride compound produced in the reaction?

Short Answer

Expert verified
The formula of the manganese chloride compound produced in the reaction is MnCl5.

Step by step solution

01

Determine the number of moles of manganese metal and hydrogen gas

We have the mass of manganese (2.747g) and can calculate the number of moles of manganese with the help of its molar mass (54.94g/mol): \(moles_{Mn} = \frac{mass_{Mn}}{Molar \: mass} = \frac{2.747g}{54.94g/mol} = 0.0500 mol\) We have the volume, temperature, and pressure of the hydrogen gas produced, and can use the Ideal Gas Law (PV=nRT) to find the moles of hydrogen gas: \(P = 0.951 \: atm\) \(V = 3.22 \: L\) \(T = 373 \: K\) \(R = 0.0821\: \frac{L \cdot atm}{mol \cdot K}\) \(n_{H_2} = \frac{PV}{RT} = \frac{(0.951atm)(3.22L)}{(0.0821\frac{L\cdot atm}{mol \cdot K})(373K)} = 0.123\: mol\)
02

Use stoichiometry to find the mole ratio between manganese and chlorine

We know the balanced reaction should look like: \(Mn + xHCl \rightarrow MnCl_x + \frac{x}{2}H_2\) From Step 1, we found the moles of Mn and H2 produced: \(0.0500 \: mol \: Mn\) \(0.123 \: mol \: H_2\) Since the mole ratio between Mn and H2 in the balanced reaction is 1: \(\frac{x}{2}\), we have: \(\frac{0.0500}{1} = \frac{0.123}{\frac{x}{2}} \Rightarrow x = 4.94\) Now, since x must be a whole number, we round it to the nearest integer, which is 5.
03

Determine the formula of the manganese chloride compound

Now that we know the value of x, we can determine the formula of the MnClx compound: The formula of the manganese chloride compound is MnCl5.

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Most popular questions from this chapter

One way of separating oxygen isotopes is by gaseous diffusion of carbon monoxide. The gaseous diffusion process behaves like an effusion process. Calculate the relative rates of effusion of $^{12} \mathrm{C}^{16} \mathrm{O},^{12} \mathrm{C}^{17} \mathrm{O},\( and \)^{12} \mathrm{C}^{18} \mathrm{O} .$ Name some advantages and disadvantages of separating oxygen isotopes by gaseous diffusion of carbon dioxide instead of carbon monoxide.

In the "Methode Champenoise," grape juice is fermented in a wine bottle to produce sparkling wine. The reaction is $$\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(a q) \longrightarrow 2 \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(a q)+2 \mathrm{CO}_{2}(g)$$ Fermentation of \(750 .\) mL grape juice (density $=1.0 \mathrm{g} / \mathrm{cm}^{3} )$ is allowed to take place in a bottle with a total volume of 825 \(\mathrm{mL}\) until 12\(\%\) by volume is ethanol $\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\( . Assuming that the \)\mathrm{CO}_{2}$ is insoluble in \(\mathrm{H}_{2} \mathrm{O}\) (actually, a wrong assumption), what would be the pressure of \(\mathrm{CO}_{2}\) inside the wine bottle at \(25^{\circ} \mathrm{C} ?\) (The density of ethanol is $0.79 \mathrm{g} / \mathrm{cm}^{3} . )$

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Do all the molecules in a 1 -mole sample of \(\mathrm{CH}_{4}(g)\) have the same kinetic energy at 273 \(\mathrm{K}\)? Do all molecules in a 1 -mole sample of \(\mathrm{N}_{2}(g)\) have the same velocity at 546 \(\mathrm{K}\) ? Explain.

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