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

Respuesta :

i think the answer is a. 91.29grams

Answer: 91.29 grams

Explanation:

[tex]2Al_2O_3(s)+3352kJ\rightarrow 4Al(s)+3O_2(g)[/tex]

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 [tex]Al_2O_3[/tex]

Thus 1500 kJ of energy will be absorbed by=[tex]\frac{2}{3352}\times 1500=0.89moles[/tex] of [tex]Al_2O_3[/tex]

Mass of [tex]Al_2O_3=moles\times {\text {Molar mass}}=0.89\times 102g/mol=91.29g[/tex]