Question

How many grams of Boron are produced when 21.77g of B2O3 are allowed to react with 27.52g of Mg?

B2O3 + 3Mg --> 2B + 3MgO

Answer 1

Answer – We are given, mass of B₂O₃ = 21.77 g , mass of Mg = 27.52 g

Reaction - B₂O₃ + 3Mg -----> 2B + 3MgO

We need to calculate moles of both reactant fist

Moles of B₂O₃ = 21.77 g / 69.62 g.mol^-1 = 0.313 moles

Moles of Mg = 27.52 g / 24.305 g.mol^-1 = 1.13 moles

Now we need to determine the limiting reactant-

Moles of B from B₂O₃

From the above balanced reaction

1 moles of B₂O₃ = 2 moles of B

So, 0.313 moles of B₂O₃ = ?

= 0.625 moles of B

Moles of B from Mg

From the above balanced reaction

3 moles of Mg = 2 moles of B

So, 1.13 moles of Mg= ?

= 0.755 of B

So the moles of B is lowest form the B₂O₃ , so limiting reactant is B₂O₃ and

Moles of B = 0.625 moles

Mass of B = 0.625 moles * 10.811 g/mol

                  = 6.76 g

So, 6.76 grams of Boron are produced when 21.77g of B2O3 are allowed to react with 27.52g of Mg