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Consider the balanced reaction below:

4Al+302=2Al203

How many grams of oxygen is needed to completely react with 9.30 moles of aluminum?

[ ] g O2

Respuesta :

znk

The mass needed is [223] g O₂.  

M_r:    26.98    32.00  

             4Al   +   3O₂ ⟶ 2Al₂O₃

n/mol:  9.30

Moles of O₂ = 9.30 mol Al × (3 mol O₂/4 mol Al) = 6.975 mol O₂

Mass of O₂= 6.975 mol O₂× (32.00 g O₂/1 mol O₂) = 223 g O₂

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