4.10 CYL - Chemistry (OpenStax)

Question

What mass of gallium oxide Ga₂O₃ can be prepared from 29 g of gallium metal? The equation for the reaction is  \(4Ga + 3{O_2} \to 2G{a_2}{O_3}\)

Step-by-Step Solution

Answer

39g of Gallium oxide can be prepared from 29 g of gallium metal.

Calculating the number of moles of gallium oxide

Number of moles in 29 g of gallium metal 

\(\begin{array}{l} = \frac{{29}}{{70\left( {rounding\,off\,69.7} \right)}}\\ = 0.414\,mol\end{array}\)

From the balanced reaction, we can say that 2 moles of gallium will produce 1 mole of gallium oxide, So, 0.414 moles of gallium will produce 0.207 mol of gallium oxide.

Calculating the mass of gallium oxide

Using the formula of moles, we get

Moles = mass/molecular mass

\(\begin{array}{c}0.207 = \frac{x}{{187.4}}\\x = 38.7\end{array}\)

Therefore, 39g of Gallium oxide can be prepared.