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Chapter 9: Problem 48

An intermetallic compound is found in the aluminum-zirconium system that has a composition of \(22.8 \mathrm{wt} \%\) Al-77.2 wt \(\% \mathrm{Zr}\). Specify the formula for this compound.

Short Answer

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Question: Determine the empirical formula of the intermetallic compound in the aluminum-zirconium system, where the weight percentages are 22.8% Al and 77.2% Zr.

Step by step solution

01

Calculate the moles of Al and Zr

To calculate the moles of Al and Zr in the compound, we can use the formula: \( moles = \frac{weight}{molar\ weight}\) Using the weight percentages given, we can determine the amount of Al and Zr in 100 grams of the compound: \( weight_{Al} = 22.8\ g\) \( weight_{Zr} = 77.2\ g\) Now, we can find the moles of Al and Zr using their respective molar weights (Al = 26.98 g/mol, Zr = 91.22 g/mol): \( moles_{Al} = \frac{22.8}{26.98}\) \( moles_{Zr} = \frac{77.2}{91.22}\)
02

Determine the mole ratio of Al and Zr

To determine the mole ratio between Al and Zr, we can simply divide their individual moles. First, we will determine the smallest number of moles: \( min\ moles = min \left( moles_{Al}, moles_{Zr} \right) \) Then, calculating the mole ratio: \( mole\ ratio_{Al} = \frac{moles_{Al}}{min\ moles}\) \( mole\ ratio_{Zr} = \frac{moles_{Zr}}{min\ moles}\)
03

Determine the empirical formula of the compound

Based on the mole ratio obtained in step 2, we can determine the intermetallic compound's empirical formula: $$ Al_{mole\ ratio_{Al}}Zr_{mole\ ratio_{Zr}} $$ Round the mole ratios to the nearest whole number to obtain the empirical formula of the compound. Note that, in some cases, the mole ratios may not be whole numbers. If this happens, try multiplying both ratios by a constant number (2,3,4, etc.) until both values are whole numbers.

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Most popular questions from this chapter

For a series of \(\mathrm{Fe}-\mathrm{Fe}_{3} \mathrm{C}\) alloys with compositions ranging between \(0.022\) and \(0.76 \mathrm{wt} \% \mathrm{C}\) that have been cooled slowly from \(1000^{\circ} \mathrm{C}\), plot the following: (a) mass fractions of proeutectoid ferrite and pearlite versus carbon concentration at \(725^{\circ} \mathrm{C}\) (b) mass fractions of ferrite and cementite versus carbon concentration at \(725^{\circ} \mathrm{C}\).

Is it possible to have an iron-carbon alloy for which the mass fractions of total cementite and proeutectoid ferrite are \(0.057\) and \(0.36\), respectively? Why or why not?

The microstructure of an iron-carbon alloy consists of proeutectoid ferrite and pearlite; the mass fractions of these two microconstituents are \(0.174\) and \(0.826\), respectively. Determine the concentration of carbon in this alloy.

Consider \(1.5 \mathrm{~kg}\) of a \(99.7 \mathrm{wt} \% \mathrm{Fe}-0.3 \mathrm{wt} \%\) C alloy that is cooled to a temperature just below the eutectoid. (a) How many kilograms of proeutectoid ferrite form? (b) How many kilograms of eutectoid ferrite form? (c) How many kilograms of cementite form?

For \(5.7 \mathrm{~kg}\) of a magnesium-lead alloy of composition \(50 \mathrm{wt} \% \mathrm{~Pb}-50 \mathrm{wt} \% \mathrm{Mg}\), is it possible, at equilibrium, to have \(\alpha\) and \(\mathrm{Mg}_{2} \mathrm{~Pb}\) phases with respective masses of \(5.13\) and \(0.57 \mathrm{~kg} ?\) If so, what will be the approximate temperature of the alloy? If such an alloy is not possible, then explain why.
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