Question

If you start with 17.2 g of Al2(SO3)3 and you produce 6.83 gAl(OH)3, what is your percent yield for this reaction?

Al2(SO3)3 + 6 NaOH → 3 Na2SO3 + 2 Al(OH)3

Answer 1

`Y=75\%`

Step-by-step explanation:

Hello,

In this case, since the aluminum sulfite, whose molar mass is 294.153 g/mol, is in a 1:2 molar ratio with aluminum hydroxide, whose molar mass is 78 g/mol, we can perform the following stoichiometric procedure to compute the theoretical yield of aluminum hydroxide to furtherly compute the percent yield:

`m_(Al(OH)_3)=17.2gAl_2(SO_3)_3*(1molAl_2(SO_3)_3)/(294.153gAl_2(SO_3)_3) *(2molAl(OH)_3)/(1molAl_2(SO_3)_3) *(78gAl(OH)_3)/(1molAl(OH)_3) \\\\m_(Al(OH)_3)=9.12gAl(OH)_3`

So we compute the percent yield considering the actual yield is 6.83 g:

`Y=(m^(actual))/(m^(theoretical)) *100\%=(6.83g)/(9.12g) *100\%\\\\Y=75\%`

Regards.