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

Short Answer

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

Step by step solution

01

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.

02

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.

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