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?
The balanced equation I came up with is this: B2O3 + 3Mg --> 2B + 3MgO
Then I converted the 125 kg of Mg to moles, and I got 5141.917 moles. I also converted the boron oxide to moles, and I got 2125.487 moles. I think that Mg is the limiting reactant, because in the balanced equation 3 moles of Mg is needed for every mole of boron oxide, which means that the 2125.487 moles of boron oxide would require 6376.567 moles of Mg. Therefore, we wouldn't have enough Mg and this would be the limiting reactant. However, I don't know how to proceed from here, and figure out what mass of boron can be produced.
M7
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Re: M7
You would first start off by converting the 125kg of B2O3 to grams so you should get 125x10^3g B2O3. You would then multiply that value by 1 mole of B2O3/ 69.619g B2O3, times 2 mol B/1mol B2O3 (gotten from the balanced equation), times 10.811 g B/1 mol B. You should get the answer 3.88x10^4 g B produced from B203.
You would use the same process to know how much boron is produced from Mg, but changing the values of g/mol for Mg as well as using the stoichiometric coefficients from the balanced equation that correspond with the proper elements being used (Mg and B). You would then know which is the limiting reactant (should be Mg since it produced 3.71x10^4 g B).
Hope this helps!
You would use the same process to know how much boron is produced from Mg, but changing the values of g/mol for Mg as well as using the stoichiometric coefficients from the balanced equation that correspond with the proper elements being used (Mg and B). You would then know which is the limiting reactant (should be Mg since it produced 3.71x10^4 g B).
Hope this helps!
Re: M7
Did we need to achieve the mass in grams for B for both B203 and Mg to find the limiting reactant?
Couldn't we just stop after we found the moles of B each we produce, determine the limiting factor then?
Or did you just go all thee way to grams since the answer asked for the mass of B in grams.
Couldn't we just stop after we found the moles of B each we produce, determine the limiting factor then?
Or did you just go all thee way to grams since the answer asked for the mass of B in grams.
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