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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

Respuesta :

sebassandin

Answer:

[tex]Y=75\%[/tex]

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:

[tex]m_{Al(OH)_3}=17.2gAl_2(SO_3)_3*\frac{1molAl_2(SO_3)_3}{294.153gAl_2(SO_3)_3} *\frac{2molAl(OH)_3}{1molAl_2(SO_3)_3} *\frac{78gAl(OH)_3}{1molAl(OH)_3} \\\\m_{Al(OH)_3}=9.12gAl(OH)_3[/tex]

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

[tex]Y=\frac{m^{actual}}{m^{theoretical}} *100\%=\frac{6.83g}{9.12g} *100\%\\\\Y=75\%[/tex]

Regards.

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