How many grams of oxygen gas (02) are needed to completely react with 9.30 moles
of aluminum?

4A1+302 → 2Al2O3

9.30 mol Al
3 mol O₂
4 mol Al

32.0 g 0₂
1 mol O₂

[?] g 0₂

Respuesta :

Answer:

223 g O₂

Explanation:

To find the mass of oxygen gas needed, you need to (1) convert moles Al to moles O₂ (via the mole-to-mole ratio from reaction coefficients) and then (2) convert moles O₂ to grams O₂ (via the molar mass). When writing your ratios/conversions, the desired unit should be in the numerator in order to allow for the cancellation of the previous unit. The final answer should have 3 sig figs because the given value (9.30 moles) has 3 sig figs.

4 Al + 3 O₂ ----> 2 Al₂O₃
^         ^

Molar Mass (O₂): 32.0 g/mol

9.3 moles Al          3 moles O₂              32.0 g
-------------------  x  ---------------------  x  --------------------  =  223 g O₂
                              4 moles Al               1 mole