Textbook M7

[Grace Chen 3F]

M.7 Solid boron can be extracted from solid boron oxide by reaction with magnesium metal at a high temperature. A second product is solid magnesium oxide. (a) Write a balanced equation for the reaction. (b) What mass of boron can be produced when 125 kg of boron oxide is heated with 125 kg of magnesium?

For part (b) of this question, I found out that the limiting reactant is 125kg of Mg. However, in the solution manual, the answer used 125kg of boron oxide (B2O3) as limiting reactant. I am really confused, can anyone explain to me why boron oxide is the limiting reactant?

[Antonia Valencia 2H]

Hi! You are correct in believing that the limiting reactant is the 125 g of Magnesium. Moreover, when I look at the solutions manual, the textbook does actually utilize Mg as the limiting reactant, and in consequence, by using this limiting reactant, you can produce 3.71 x 10^4 grams of Boron. I hope this helps your confusion!

[Rose Arcallana 2B]

So first you have to do balance the equations to get the mole ratios. Next, convert your masses into moles. To do that you divide the mass by the molar mass of each compound. So for B2O3, we get 1795.41 moles and for Mg, we get 5142.97 moles. To determine the limiting reactants, you take your moles and multiply them by mole ratios. So to test B2O3, you take its moles and multiply by 3 (stoichiometry coefficient of Mg) / 1 (stoichiometry coefficient of B2O3) = 5386.23 > 5142.97 original mole of Mg, this means B2O3 CANNOT be the limiting reactant because you would need more than you have. So for Mg, you do the same, you take its moles and multiply by 1 (stoichiometry coefficient of B2O3) / 3 (stoichiometry coefficient of Mg) = 1714.32 < 1795.41 original mole of B2O3, so Mg is our limiting Reactant.

To determine the product of B, the same concept as before. Use the limiting reactant mole multiply by 2 (stoichiometry coefficient of B) / 3 (stoichiometry coefficient of Mg) = 3428.65 moles. To find the mass multiply by the molar mass of B, this should give you your final answer. :)