Aluminum is mined as the mineral bauxite, which consists...

Aluminum is mined as the mineral bauxite, which consists primarily of $\mathrm{Al}_{2} \mathrm{O}_{3}$ (alumina). a. How much aluminum is produced from 1 metric ton. $$\begin{array}{r}\left(1 \text { metric ton }=10^{3} \mathrm{kg}\right) \text { of } \mathrm{Al}_{2} \mathrm{O}_{3} ? \\2 \mathrm{Al}_{2} \mathrm{O}_{3}(s) \rightarrow 4 \mathrm{Al}(s)+3 \mathrm{O}_{2}(g) \end{array}$$.b. The oxygen produced in part (a) is allowed to react with carbon to produce carbon monoxide:$$\mathrm{O}_{2}(g)+2 \mathrm{C}(s) \rightarrow 2 \mathrm{CO}(g)$$.Balance the following equation describing the reaction of alumina with carbon:$$\mathrm{Al}_{2} \mathrm{O}_{3}(s)+\mathrm{C}(s) \rightarrow \mathrm{Al}(s)+\mathrm{CO}(g)$$. c. How much CO can be produced from the $\mathrm{O}_{2}$ made in part (a)?

Aluminum is mined as the mineral bauxite, which consists primarily of $\mathrm{Al}_{2} \mathrm{O}_{3}$ (alumina).
a. How much aluminum is produced from 1 metric ton.
$$\begin{array}{r}\left(1 \text { metric ton }=10^{3} \mathrm{kg}\right) \text { of } \mathrm{Al}_{2} \mathrm{O}_{3} ? \\2 \mathrm{Al}_{2} \mathrm{O}_{3}(s) \rightarrow 4 \mathrm{Al}(s)+3 \mathrm{O}_{2}(g)
\end{array}$$.b. The oxygen produced in part (a) is allowed to react with carbon to produce carbon monoxide:$$\mathrm{O}_{2}(g)+2 \mathrm{C}(s) \rightarrow 2 \mathrm{CO}(g)$$.Balance the following equation describing the reaction of alumina with carbon:$$\mathrm{Al}_{2} \mathrm{O}_{3}(s)+\mathrm{C}(s) \rightarrow \mathrm{Al}(s)+\mathrm{CO}(g)$$.
c. How much CO can be produced from the $\mathrm{O}_{2}$ made in part (a)?

Key Concepts

Molar Mass

Molar mass is the mass of one mole of a substance and is used to convert between the mass of a substance and the number of moles. This conversion is crucial for applying stoichiometric ratios from balanced equations to determine how much product forms from a given mass of reactant.

Stoichiometry

Stoichiometry is the study of the quantitative relationship between reactants and products in a chemical reaction. It involves calculations based on the mole ratios from balanced chemical equations to determine how much product is obtained from given amounts of reactants.

Mole Concept

The mole concept provides a bridge between the atomic scale and laboratory-scale amounts of substances. It defines the mole as a specific number of particles (Avogadro's number) and allows for conversion between mass and number of entities through the molar mass, which is essential for quantitative reaction analysis.

Balanced Chemical Equations

Balanced chemical equations are fundamental to ensuring the conservation of mass in chemical reactions. They provide the necessary stoichiometric coefficients that represent the proportions of reactants and products, enabling accurate stoichiometric calculations and predictions of reaction outcomes.

Transcript

00:01 For problem 122, we have to convert the kilogram of aluminum oxide to gram because in the modern mass, we use gram, we don't give you kilogram.

00:13 Okay, so let's see.

00:17 10, 1, 2, 2.

00:32 And then from this, we use to, from this gram, the amount of this gram, so we go to find the aluminum in this equation question a first oh hold on this we're aluminum right and then we get oxygen and now we have 10 power 6 gram aluminum oxide the strategy we do similar to to reverse problem gram to more a more a to more b and more b to gram of b and how many.

02:11 So you will find here aluminum equal 5 .29 time 10 b or 5 gram aluminum and if you convert to kilogram you get 5 .29 tight 10 by 2 kilograms okay now for question b we're going to answer question b and we balance this equation then c.

03:02 C you have to find the more of co oxygen in question a first and then you use that amount to find the co in question the question c plus co2, c plus oxygen to reduce co okay not right number more by the same we have 10 power 6 gram aluminum oxide, one more aluminum oxide, divided by 1 or 2 gram, aluminum oxide.

04:04 And then we tie how many more, 3 more of osses, 3 more of oxygen, and 2 more aluminum oxide.

04:38 And then you will have, let me see how many more here.

04:43 And now this is the amount of oxygen in a.

05:24 And from this amount you go to find the oxygen in this equation.

05:32 Okay, so we can show here first.

05:52 So you need more oxygen here.

05:55 And we have c, that oxygen you produce co...

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