How many grams of Boron can be obtained from 234 grams of B2O3?

72.67g of B

Explanation:

The reaction of B₂O₃ to produce boron (B), is:

B₂O₃ → 3/2O₂ + 2B

That means B₂O₃ produce 2 moles of boron

Molar mass of B₂O₃ is 69.62g/mol. 234g of B₂O₃ contains:

234g B₂O₃ ₓ (1mol / 69.62g) = 3.361 moles of B₂O₃.

As 1 mole of B₂O₃ produce 2 moles of B, Moles of B that can be produced from B₂O₃ is:

3.361mol B₂O₃ ₓ 2 = 6.722 moles of B.

As molar mass of B is 10.811g/mol. Thus mass of B that can be produced is:

6.722mol B ₓ (10.811g / mol) = 72.67g of B

the density of gold = 19.3 g/cm3

the relation between density and mass and volume is

density = mass / volume

given :

mass of gold = 13 g

density = 13g / volume = 19.3

volume = 13 /19.3 = 0.67 cm^3

thus volume of gold nugget = 0.67 cm^3