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Ivahew [28]
3 years ago
10

A student determines the aluminum content of a solution by first precipitating it as aluminum hydroxide, and then decomposing th

e 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?
Chemistry
1 answer:
Vladimir [108]3 years ago
4 0

Answer: The student should obtain <u>1.103 g of aluminum oxide </u>

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

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