The rock in a particular iron ore deposit contains 76% Fe2O3 by mass. Help please! How would I solve this problem?
Its better then the rest, there's no extra chambers, its alot smoother air flow, you're gas milleage should go up.
Each kg of rock contains, on the average, 760 grams of Fe2O3. The atomic weight of iron is 55.84, while that of oxygen is 16.00. So the actual iron in the 760 grams of Fe2O3 is 760 g times [(2*55.84) / (2*55.84 + 3*16.00) ] 531.54 grams Hence, to obtain 2200 kg of iron, the minimum mass of rock you need is 2200 kg/0.53154 4139 kg
I've done this to two car's, both Japanese, and both times was advised to use Pipercross.
