Question

How many grams of Al2O3 are required to react completely with 1500.0 kJ of heat? 2Al2O3(s) + 3352 kJ → 4Al(s) + 3O2(g)

answers
91.29 grams

2 grams

45.64 grams

1500 grams

Answer 1

i think the answer is a. 91.29grams

Answer 2

Answer: 91.29 grams

Step-by-step explanation:

`2Al_2O_3(s)+3352kJ\rightarrow 4Al(s)+3O_2(g)`

Endothermic reactions are those chemical reactions in which heat is absorbed by the reactants.

According to stoichiometry,

3352 kJ of energy is absorbed by 2 moles of
`Al_2O_3`

Thus 1500 kJ of energy will be absorbed by=
`(2)/(3352)* 1500=0.89moles` of
`Al_2O_3`

Mass of
`Al_2O_3=moles* {\text {Molar mass}}=0.89* 102g/mol=91.29g`