Question

of O, reacted with ne Na? (b) What mass of Na reacted? (c) What mass of Na,O formed? When aluminum metal is exposed to oxygen coating of aluminum oxide forms on the surf aluminum. The balanced equation for the reaction of aluminum metal with oxygen gas is 22 Limitin 6.29 4Als) +302(g) 2ALO3(s) Suppose a sheet of pere aluminum gains 0.0900 g of mass when exposed tn air. Assume that this gain can be attributed to its reactio with oxygen. (a) What mass of O2 reacted with the Al (b) What mass of Al reacted? (c) What mass of Al,0, formed? Diiodine pentoxide is used in respirators to remo 6.30 6.31 monoxide from air:

Answer 1

Ans 22 :

a) Since the aluminium sheet gained the mass =0.0900 g , so the mass of oxygen reacted will also be sam , i.e. 0.0900 g.

b) Number of moles of oxygen = 0.0900 / 31.99 = 0.0028 moles

3 moles of O2 requires 4 moles of Al

So 0.0028 moles of it will require ( 0.0028 x 4) / 3

= 0.0037 moles

Mass of Aluminium reacted = 0.0037 x 26.98

= 0.099 grams

c) The mass of Al2O3 formed will be = 0.0900 + 0.099 = 0.189 grams