Question

The thermite reaction occurs when iron(III) oxide reacts with solid

Answer 1

Answer:

$m_{Al}=3.38gAl$

$Y=83.1\%$

Explanation:

Hello.

In this case, for the given balanced reaction:

$Fe_2O_3(s) + 2Al(s)\rightarrow 2Fe(s) + Al_2O_3(s)$

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:

$m_{Al}=10.0gFe_2O_3*\frac{1molFe_2O_3}{160gFe_2O_3} *\frac{2molAl}{1molFe_2O_3} *\frac{27gAl}{1molAl} \\\\m_{Al}=3.38gAl$

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:

$m_{Al_2O_3}=10.0gFe_2O_3*\frac{1molFe_2O_3}{160gFe_2O_3} *\frac{1molAl_2O_3}{1molFe_2O_3} *\frac{102gAl_2O_3}{1molAl_2O_3} \\\\m_{Al_2O_3}=6.38gAl_2O_3$

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

$Y=\frac{5.3 g}{6.38 g}* 100\%\\\\Y=83.1\%$

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