Question

help me do the correct math

25. (12 pts) Aluminum oxide forms when aluminum-reacts with oxygen. Balance the following chemical equation (the molar mass of wluminum oxide is 101.96 g/mol): 4 Al+ 3 02 2 Al2O3 62.79 g of aluminum metal is allowed to react. How many moles of aluminum is this? 52.79 gray.101.96 glmolo = 6402 moles Al How many moles of Al2O3 can be produced from the previously calculated moles of aluminum? 23 6402 x 6o2to² 3 =31858 10 27 What mass of aluminum oxide can be formed from the moles of Al2O3 just calculated? Al=2x26.98=3396 -3 O = 3x 16.0o = 48.22 101.96 101.95 g/mol

Answer 1

The reaction of Aluminum with oxygen can be written as follows:

$Al + O_2 \rightarrow Al_2O_3$

To balance the number of O atoms on both side, we cross multiply the number of oxygens. i.e. we multiply 2 on Al2O3 and multiply 3 on O2.

$Al + 3O_2 \rightarrow 2Al_2O_3$

Next, to balance the number of Al atom, we realise that there are 4 Al on right and only one on left. Hence, we multiply 4 on Al to get the balanced reaction.

${\color{Red} 4}Al + {\color{Red} 3}O_2 \rightarrow {\color{Red} 2}Al_2O_3$

Now, mass of Aluminum allowed to react with oxygen is 62.79 g.

Atomic mass of Al = 26.98 g/mol

Hence, number of moles of Aluminum in 62.79 g can be calculated as

$moles = \frac{mass}{molar \ mass} = \frac{62.79 \ g}{26.98 \ g\ mol^{-1}}\approx {\color{Red} 2.327 \ mol }$

Hence, there are 2.327 mol of Al in the given amount of Aluminum.

Note that from the balanced equation, 4 mol of Aluminum forms 2 moles of Aluminum oxide.

Hence, the given 2.327 mol of Aluminum above will be able to form:

$\frac{2 \ mol \ Al_2O_3}{4 \ \cancel{mol \ Al}} \times 2.327 \ \cancel{mol \ Al} = {\color{Red} 1.1635 \ mol \ Al_2O_3}$

Hence, 1.164 mol of Al2O3 approximately can be formed from the given amount of Al.

Molar mass of Al2O3 = 101.96 g/mol

Hence, mass of Al2O3 in 1.1635 mol of Al2O3 can be calculated as

$mass = moles \times molar \ mass = 1.1635 \ \cancel{mol} \times 101.96 \frac{g}{\cancel{mol}} \approx{\color{Red} 118.6 \ g}$

Hence, approximately 118.6 g of Al2O3 can be formed from the given amount of Al.