Question

What is the minimum number of grams of O2 gas required to produce 102 grams of Al2O3?

192 g

32.0 g

48.0 g

96.0 g

Answer 1

1 mol Al2O3 x (3 mol O2 / 2 mol AL2O3) = 1.5 mol O2

1.5 mol O2 x (32g O2 / 1 mol O2) = 48g O2

Coefficients are mole ratios in a balanced equaion.

So 48g of O2 is required to to produce 102 grams of Al2O3.

Answer 2

48.0 g $O_2$ gas required to produce 102 grams of $Al_2O_3$

Explanation:

The balanced chemical reaction for formation of $Al_2O_3$

$4Al+3O2\rightarrow 2Al_2O_3$

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$  

$\text{Number of moles of}Al_2O_3=\frac{102g}{102g/mol}=1mol$  

Based on the reaction stoichiometry:

2 moles of $Al_2O_3$ are produced by 3 moles of  $O_2$

Therefore, 1 moles of $Al_2O_3$ is produced by =$\frac{3}{2}\times 1=1.5$ moles of  $O_2$

$1.5mol=\frac{\text{Mass of oxygen}}{32g/mol}\\\\\text{Mass of oxygen}=48g$

Thus  48.0 g $O_2$ gas required to produce 102 grams of $Al_2O_3$

Learn more about stoichiometry

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