Answer:
39.06 g
Step-by-step explanation:
The balanced chemical equation for the reaction between gallium metal and oxygen to form gallium oxide is:
4 Ga + 3 O2 → 2 Ga2O3
According to the equation, 4 moles of gallium react with 3 moles of oxygen to produce 2 moles of Ga2O3.
We need to determine the mass of Ga2O3 that can be produced from 29.0 g of gallium metal.
First, we need to calculate the number of moles of Ga present in 29.0 g of Ga. The molar mass of Ga is 69.72 g/mol, so the number of moles of Ga is:
n(Ga) = m(Ga) / M(Ga)
n(Ga) = 29.0 g / 69.72 g/mol
n(Ga) = 0.4167 mol
According to the balanced equation, 4 moles of Ga produce 2 moles of Ga2O3. Therefore, the number of moles of Ga2O3 produced is:
n(Ga2O3) = n(Ga) × (2 moles Ga2O3 / 4 moles Ga)
n(Ga2O3) = 0.4167 mol × 0.5
n(Ga2O3) = 0.2084 mol
Finally, we can calculate the mass of Ga2O3 produced using its molar mass, which is 187.44 g/mol:
m(Ga2O3) = n(Ga2O3) × M(Ga2O3)
m(Ga2O3) = 0.2084 mol × 187.44 g/mol
m(Ga2O3) = 39.06 g
Therefore, the mass of gallium oxide, Ga2O3, that can be prepared from 29.0 g of gallium metal is 39.06 g.