Question

163 mL of 2.75 mol/L aluminum sulfate is to have all aluminum ions removed by adding a 3.65 mol/L sodium sulfide solution.
Al2(SO4)3(aq) + 3 Na2S(aq) → Al2S3(s) + 3 Na2SO4(aq)

What is the minimum volume of sodium sulfide that must be added to completely remove all the aluminum ions? What mass of aluminum sulfide will form?

Answer 1

Answer:

$V_{Na_2S}=368mL$

$m_{Al_2S_3}=67.3gAl_2S_3$

Explanation:

Hello there!

In this case, according to the given information, it is possible to realize that the only way for us to calculate the required volume of sodium sulfide, is by calculating the moles of this substance consumed 163 mL of 2.75 mol/L aluminum sulfate by using the definition of molar concentration and the 1:3 mole ratio between these two:

$n_{Na_2S}=0.163L*2.75\frac{molAl_2(SO_4)_3}{L}*\frac{3molNa_2S}{1molAl_2(SO_4)_3} =1.34molNa_2S$

Now, we divide these moles by the molar concentration of sodium sulfide to obtain the required volume:

$V_{Na_2S}=\frac{1.34molNa_2S}{3.65mol/L} =0.368L=368mL$

For the last part, we now use the 1:1 mole ratio of aluminum sulfate to aluminum sulfide and the molar mass of the latter (150.158 g/mol) in order to calculate the required mass:

$m_{Al_2S_3}=0.163L*2.75\frac{molAl_2(SO_4)_3}{L}*\frac{1molAl_2S_3}{1molAl_2(SO_4)_3} *\frac{150.158gAl_2S_3}{1molAl_2S_3} \\\\m_{Al_2S_3}=67.3gAl_2S_3$

Regards!