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The thermite reaction occurs when iron(III) oxide reacts with solid
aluminum. The reaction is so hot that molten iron forms as a product.
>
Fe2O3(s) + Al(s) → Fe(C) + Al2O3(s)
What mass of aluminum should be used in order to completely
consume 10.0 g Fe2O3(s)? If the reaction described produces 5.3 g
Al2O3(s), what is the percent yield?

Respuesta :

sebassandin

Answer:

[tex]m_{Al}=3.38gAl[/tex]

[tex]Y=83.1\%[/tex]

Explanation:

Hello.

In this case, for the given balanced reaction:

[tex]Fe_2O_3(s) + 2Al(s)\rightarrow 2Fe(s) + Al_2O_3(s)[/tex]

For 10.0 g of iron (III) oxide (molar mass = 160 g/mol), based on the 1:2 mole ratio with Al (atomic mass = 27 g/mol), the required mass is then:

[tex]m_{Al}=10.0gFe_2O_3*\frac{1molFe_2O_3}{160gFe_2O_3} *\frac{2molAl}{1molFe_2O_3} *\frac{27gAl}{1molAl} \\\\m_{Al}=3.38gAl[/tex]

Moreover, as 5.3 g of aluminum oxide are actually yielded, from the 10.0 g of iron (III) oxide, we can compute the theoretical mass of aluminum oxide (molar mass = 102 g/mol) via their 1:1 mole ratio:

[tex]m_{Al_2O_3}=10.0gFe_2O_3*\frac{1molFe_2O_3}{160gFe_2O_3} *\frac{1molAl_2O_3}{1molFe_2O_3} *\frac{102gAl_2O_3}{1molAl_2O_3} \\\\m_{Al_2O_3}=6.38gAl_2O_3[/tex]

Thus, the percent yield (actual/theoretical*100%) turns out:

[tex]Y=\frac{5.3 g}{6.38 g}* 100\%\\\\Y=83.1\%[/tex]

Best regards.