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Industrial Scale

Part of the Stoichiometry section of Coddy's Chemistry journey. Lesson 27 of 41.

Factories count in tons: 1 ton = 1000 kg = 1×106 g1 \times 10^{6}\ \mathrm{g}. The balanced equation and its mole ratio are the same as in a test tube.

Aluminum comes from splitting molten aluminum oxide: 2Al₂O₃ → 4Al + 3O₂. For 1 ton of Al: 1×106 g1 \times 10^{6}\ \mathrm{g} ÷ 27 = 3.70×1043.70 \times 10^{4} mol Al, so 1.85×1041.85 \times 10^{4} mol Al₂O₃.

Times 102 g/mol: 1.89×106 g1.89 \times 10^{6}\ \mathrm{g}, which is 1.89 tons of Al₂O₃ for every ton of aluminum.

Two bars: 1.89 tons of aluminum oxide is used to make 1 ton of aluminum

About 1.89 tons of Al₂O₃ for every ton of aluminum

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Challenge

A smelter must make 5.4 tons5.4\ \mathrm{tons} of aluminum by splitting molten aluminum oxide: 2Al₂O₃ → 4Al + 3O₂. What mass of Al₂O₃ does it need, in grams?

Try it yourself

Physics diagram: balance

Solve for moxm_{ox}

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This lesson includes a short quiz. Start the lesson to answer it and track your progress.

All lessons in Stoichiometry