Hi @Naydoe .. your numbers sound about right for 10,000 of ferro...

Hi @Naydoe .. your numbers sound about right for 10,000 of ferro niobium...

Using your percentages, if we start with 508,804.4485 tonnes of 'ore' at 4.15% grade, we end up with 21,115.3846 tonnes of Nb2O5 taken out of the ground. Multiply this by 0.52 to get our recovery = 10,980 tonnes recovered..

As the concentrate is 61% .. we divide the 10,980 by 0.61 = 18,000 tonnes of concentrate. With this concentrate we make 11,000 tonnes of Niobium.

In the process we lose the oxygen from niobium pentoxide but add iron to make the ferro niobium molecule. Iron at an atomic weight is around 56 is much heavier than oxygen at 16.
At FeNb for a molecule of ferro niobium, we are turning one molecule of Nb2O5 into 2 molecules of FeNb, so losing 5 atoms of oxygen or a total atomic weight of 90 (5 X 16) and gaining 2 atoms of Fe (2 X 56 = 112).


In other words we are still recovering around the 52% of the total mined as ferro niobium at the end of the process. We lose a very small amount in the processing but it's made up for by the fact 1 iron atom is heavier than 2.5 oxygen atoms.
Final product is each tonne of ferro niobium has just over 62% Niobium and just under 38% oxygen. 92/(92 + 56) at atomic level...