Phosphorite is a mineral that contains plus other non-phosphorus-containing compounds. What is the maximum amount of that can be produced from 2.3 kg of phosphorite if the phosphorite sample is 75% by mass? Assume an excess of the other reactants.

Phosphorus can be prepared from calcium phosphate by the following reaction:

Phosphorite is a mineral that contains plus other non-phosphorus-containing compounds. What is the maximum amount of that can be produced from 2.3 kg of phosphorite if the phorphorite sample is 75% by mass? Assume an excess of the other reactants.

Thus the maximum amount of that can be produced is 0.345 kg

Explanation:

Given mass of phosphorite  = 2.3 kg

As given percentage of phosphorite is =

According to stoichiometry:

2 moles of phosphorite gives = 1 mole of

Thus 5.56 moles of phosphorite give= of

Mass of

Thus the maximum amount of that can be produced is 0.345 kg

first, consult an activity series chart to determine if al is more reactive than fe. it is, so it will replace the fe. now, one can continue with the problem.

next, make sure you have a correctly balanced equation.

3feo + 2al ==> 3fe + al2o3

now, determine which reactant (feo or al) is in limiting supply.

moles feo present = 125 g x 1 mole/71.8 g = 1.74 moles

moles al present = 25.0 g x 1 mole/26.98 g = 0.927 moles

since from the balanced equation, it take 3 mol feo for every 2 mol al, al will be limiting.

you can determine this simply by dividing moles of each by their coefficient in the balanced equation. for example: 1.74/3 = 0.58 and 0.927/2 = 0.464. al is less, so it is limiting.

finally, using the stoichiometry in the balanced eq. determine mass of fe that can be produced from 0.927 moles of al.

0.927 moles al x 3 moles fe/2 moles al x 55.8 g/mole = 77.6 g

c is the answer because when something is left after something evaporates that shows the chemical change.