Astudent determines the aluminum content of a solution by first precipitating it as aluminum hydroxide, and then decomposing the hydroxide to aluminum oxide by heating. how many grams of aluminum oxide should the student obtain if her solution contains 40.0 ml of 0.541 m aluminum nitrate?

The student should obtain 1.103 g of aluminum oxide

Explanation:

Al(NO3)3 + 3NaOH → 3NaNO3 + Al(OH)3

Now,  the equation that represents the decomposition of the hydroxide to aluminum oxide by heating it.

2Al(OH)3 → Al2O3 + 3H2O

Volumen of  Al(NO3)3 = 40mL

Molar concentration of Al(NO3)3 =  0.541M

Molecular Weight Al2O3 = 101.96 g/mol

then we use the molar coefficients from the  equations to discover the amount of Al2O3  moles produced

finally, we use the molecular weight of the Al2O3  to calculate the final mass produced.

the potential energy of the reactants is 200j.

from the energy diagram, the energy of the product formed is 350j; this means that, this reaction is an endothermic reaction, because it absorbs energy from its environment.