Question

A student 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?

Answer 1

Answer: The student should obtain 1.103 g of aluminum oxide

Explanation:

• First we write down the equations that represent the aluminum hydroxide precipitation  from the reaction between the aluminum nitrate and the sodium hydroxide:

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

• Second, we gather the information what we are going to use in our calculations.

Volumen of  Al(NO3)3 = 40mL

Molar concentration of Al(NO3)3 =  0.541M

Molecular Weight Al2O3 = 101.96 g/mol

• Third, we start using the molar concentration of the aluminum nitrate and volume used to find out the total amount of moles that are reacting

$\frac{0.541moles Al(NO3)3}{1L} x\frac{1L}{1000mL} x 40mL Al(NO3)3 =  0.022moles Al(NO3)3$

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

$0.022moles Al(NO3)3 x \frac{1mol Al(OH)3}{1mol Al(NO3)3} X\frac{1mol Al2O3}{2molAl(OH)3} = 0.011 moles Al2O3$

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

$0.011moles Al2O3 x \frac{101.96g Al2O3}{1mol Al2O3} = 1.103g Al2O3$