The semiconductor gallium arsenide, GaAs, is used in high-speed integrated circuits, light-emitting diodes, and solar cells. Its density is 5.32 \(\mathrm{g} / \mathrm{m}^{3} .\) It can be made by reacting trimethylgallium, \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{Ga},\) with arsine gas, \(\mathrm{AsH}_{3}\) . The other product of the reaction is methane, \(\mathrm{CH}_{4}\) . (a) If you reacted 450.0 g of trimethylgallium with 300.0 \(\mathrm{g}\) of arsine, what mass of GaAs could you make? (b) Which reactant, if any, would be left over, and how many moles of the leftover reactant would remain? (c) One application of GaAs uses it as a thin film. If you take the mass of GaAs from part (a) and make a 40 -nm thin film from it, what area, in \(\mathrm{cm}^{2},\) would it cover? Recall that \(1 \mathrm{nm}=1 \times 10^{-9} \mathrm{m} .\)

Step by step solution

01

Write the balanced chemical equation

We have the following chemical reaction: Trimethylgallium + Arsine → Gallium Arsenide + Methane In chemical form: \((CH_3)_3Ga + AsH_3 \rightarrow GaAs + CH_4\) When we balance the equation, we get: \((CH_3)_3Ga + AsH_3 \rightarrow GaAs + 3CH_4\)

02

Convert masses to moles

Convert the given masses of trimethylgallium (450.0 g) and arsine gas (300.0 g) into moles using their molar masses (144.65 g/mol for trimethylgallium and 77.95 g/mol for arsine). Moles of trimethylgallium = \(\frac{450.0\,g}{144.65\,g/mol} = 3.111\,mol\) Moles of arsine = \(\frac{300.0\,g}{77.95\,g/mol} = 3.849\,mol\)

03

Identify the limiting and excess reactants

Determine the mole ratio between trimethylgallium and arsine in the balanced equation (1:1). R = \(\frac{3.111\,mol}{3.849\,mol}\) = 0.808 Since R < 1, trimethylgallium is the limiting reactant, and arsine is the excess reactant.

04

Calculate the mass of GaAs produced

Use stoichiometry to determine the amount of GaAs produced from the limiting reactant (trimethylgallium): Moles of GaAs produced = Moles of trimethylgallium = 3.111 mol Now, convert moles of GaAs to mass using its molar mass (144.64 g/mol): Mass of GaAs = \(3.111\,mol \times 144.64\,g/mol = 449.9\,g\)

05

Calculate the moles of leftover reactant

Calculate the moles of leftover arsine using stoichiometry: Moles of arsine consumed = Moles of GaAs produced = 3.111 mol Moles of leftover arsine = Moles of arsine - Moles of arsine consumed = 3.849 mol - 3.111 mol = 0.738 mol

06

Calculate the area of 40-nm thin film

Determine the volume of GaAs produced using its mass (449.9 g) and density (5.32 g/cm³): Volume of GaAs = \(\frac{449.9\,g}{5.32\,g/cm^3} = 84.58\,cm^3\) Now, calculate the area of the 40-nm thin film: Thickness of the film = 40 nm = \(40 \times 10^{-9}\,m = 4 \times 10^{-7}\,cm\) Area of the thin film = \(\frac{Volume\,of\, GaAs}{Thickness\, of\, the\, film}\) Area = \(\frac{84.58\,cm^3}{4 \times 10^{-7}\, cm}\) = 2.115 x 10^8 \({cm^2}\) #Answer#: (a) The mass of GaAs that could be made is 449.9 g. (b) Arsine gas would be left over, and 0.738 moles of it would remain. (c) The area covered by the 40-nm thin GaAs film would be 2.115 x 10^8 \({cm^2}\).

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