When 81.0 g of aluminium is allowed to react with 128.0 g of oxygen gas, the mass of aluminium oxide produced in grams is ________ . (Nearest integer)

Given :

Molar mass of Al is $27.0 \mathrm{~g} \mathrm{~mol}^{-1}$

Molar mass of O is $16.0 \mathrm{~g} \mathrm{~mol}^{-1}$

<p>To determine the mass of aluminum oxide produced, we first consider the chemical reaction between aluminum and oxygen:</p> <p>$ 4 \text{Al} + 3 \text{O}_2 \rightarrow 2 \text{Al}_2\text{O}_3 $</p> <p>Given:</p> <p><p>Molar mass of Al = 27.0 g/mol</p></p> <p><p>Molar mass of O = 16.0 g/mol</p></p> <p><p>81.0 g of Al and 128.0 g of O$_2$</p></p> <p><strong>Calculating moles of reactants:</strong></p> <p>For aluminum (Al), using its molar mass:</p> <p>$ \frac{81 \text{ g}}{27 \text{ g/mol}} = 3 \text{ moles of Al} $</p> <p>For oxygen (O$_2$), using its molar mass (O$_2$ is composed of two oxygen atoms):</p> <p>$ \frac{128 \text{ g}}{32 \text{ g/mol}} = 4 \text{ moles of O}_2 $</p> <p><strong>Identifying the limiting reagent:</strong></p> <p>From the stoichiometry of the reaction:</p> <p><p>4 moles of Al react with 3 moles of O$_2$</p></p> <p><p>3 moles of Al require $ \frac{3}{4} \times 3 $ moles of O$_2$ = 2.25 moles of O$_2$</p></p> <p>Since only 2.25 moles of O$_2$ are needed by 3 moles of Al, and we have 4 moles of O$_2$, aluminum is the limiting reagent.</p> <p><strong>Calculating moles of $\text{Al}_2\text{O}_3$ formed:</strong></p> <p>From the reaction, 4 moles of Al yield 2 moles of $\text{Al}_2\text{O}_3$. Thus:</p> <p>$ \frac{3}{4} \times 2 = \frac{3}{2} \text{ moles of } \text{Al}_2\text{O}_3 $</p> <p><strong>Calculating mass of $\text{Al}_2\text{O}_3$:</strong></p> <p>The molar mass of $\text{Al}_2\text{O}_3$ is:</p> <p>$ (2 \times 27) + (3 \times 16) = 102 \text{ g/mol} $</p> <p>So, the mass of $\frac{3}{2}$ moles of $\text{Al}_2\text{O}_3$ is:</p> <p>$ \frac{3}{2} \times 102 = 153 \text{ grams} $</p> <p>Thus, the mass of aluminum oxide produced is 153 grams.</p>