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Comparing Calculated Moles step M7(b)

Posted: Thu Oct 04, 2018 11:39 pm
by Sarah Bui 2L
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?


In this question, I am confused on which ratios or required moles to use from the balanced equation to determine the limiting reactant?

Re: Comparing Calculated Moles step M7(b)

Posted: Fri Oct 05, 2018 12:03 am
by yaosamantha4F
If you go ahead and balance the chemical equation first, you get B2O3 + 3Mg(s) -> 3MgO(s) + 2B(s). You then need to calculate the mass of boron that is produced from both Mg and B2O3. First, convert the masses of both from kg to g, divide by the molar mass of the compound, multiply by the ratio of B:B2O3 from the equation (2:1), and multiply by the molar mass of boron. Do the same thing for Mg except with the molar mass of Mg instead of B2O3 and change the ratio to represent B:Mg (2:3), and compare the two. Whichever answer produces a smaller mass of B is the limiting reactant for the reaction.

Re: Comparing Calculated Moles step M7(b)

Posted: Fri Oct 05, 2018 9:18 pm
by Angel Chen 2k
Since the question is asking for the mass of boron, you calculate the mass of Boron produced from 125kg of B2O3 and 125kg of Mg. You use 2:1 ad the ratio for calculating the mass of boron produced from B2O3 and the ratio you use for the mass of boron produced from Mg is 2:3. These ratios come from the balanced chemical reaction you did in part a.