Answer:
556.88 g.
Step-by-step explanation:
- Firstly, we need to write the balanced chemical reaction:
(CH₃)₃Ga + AsH₃ → GaAs + 3 CH₄
- It is now a stichiometric oriented problem; 1.0 mole of trimethyl gallium ((CH₃)₃Ga) reacts with 1.0 mole of arsine (AsH₃) to produce 1.0 mole of gallium arsenide (GaAs) and 3.0 moles of methane (CH₄).
- Now, we need to calculate the number of moles (n) of the reacting 450.0 g of trimethyl gallium with 300.0 g of arsine using the relation:
n = mass / molar mass,
- Molar mass of trimethyl gallium ((CH₃)₃Ga) = 114.827 g/mol and molar mass of arsine (AsH₃) = 77.95 g/mole.
- n of trimethyl gallium ((CH₃)₃Ga) = mass / molar mass = (450.0 g) / (114.827 g/mole) = 3.92 mole.
- n of arsine (AsH₃) = mass / molar mass = (300.0 g) / (77.95 g/mole) = 3.85 mole.
- The limiting reactant here is arsine (AsH₃) that it has the lower number of moles.
- From the stichiometry; 1.0 mole of trimethyl gallium ((CH₃)₃Ga) reacts with 1.0 mole of arsine (AsH₃) to produce 1.0 mole of gallium arsenide GaAs.
- Thus, 3.85 mole of arsine (AsH₃) reaacts with 3.92 mole of trimethyl gallium ((CH₃)₃Ga) (which is slightly excess) to produce 3.85 mole of gallium arsenide (GaAs).
- Now, we can calculate the mass of the produced gallium arsinate from the relation: mass = n x molar mass.
- Molar mass of gallium arsenide (GaAs) = 144.645 g/mol.
- Mass of gallium arsenide (GaAs) = (3.85 mole) (144.645 g/mol) = 556.88 g.