Answer:
a. Fe(OH)2
b. H3PO4
c. 0.01185 moles
Explanation:
The balanced equation of the question will be:
3 Fe(OH)2 + 2 H3PO4 → Fe3(PO4)2 + 6 H2O
a. what is the limiting reactant?
A limiting reactant is a reactant that will be used up in the reaction. We can find it by calculating the stoichiometric ratio. The reactant with the lowest ratio, which will be depleted first, is the limiting reactant. The reaction will stop if any reactant depleted, so the limiting reactant will be the one that determines how many products made. If you add more of a limiting reactant, the reaction will continue.
The molecular mass of Fe(OH)2 is 90g/mol while the molecular mass of H3PO4 is 98 g/mol. The number of each reactant in moles will be
Fe(OH)2 = 3.20g/ (90g/mol) =0.03555 mol
H3PO4 = 2.50g/ (98g/mol) = 0.02551 mol
The actual ratio of Fe(OH)2 : H3PO4 will be:
0.03555 mol/ 0.02551 mol= 1.393 mol/ 1 mol
The stoichiometric coefficient of Fe(OH)2 is 3 while H3PO4 is 2. Mean you need 3 moles Fe(OH)2 of to for the reaction every 2 moles of H3PO4.
The stoichiometric ratio will be: 3 mol/ 2 mol= 1.5 mol/ 1 mol
Since the actual ratio (1.393) is lower than the stoichiometric ratio(1.5) that means we have less Fe(OH)2 than H3PO4, thus the limiting reactant is the Fe(OH)2.
b. What is the reactant in excess?
The excess reactant is the reactant that will have to remain after the reaction stop. It's the opposite of the limiting reactant. Since we still have remains, adding excess reactant won't continue the reaction.
We already know that Fe(OH)2 is the limiting reactant, then H3PO4 will be the excess reactant. The number of reactant remaining will be:0.02551 mol -0.03555 mol/ (1.5 mol -1 mol)=0.00181 mol
c. What is the theoretical yield
Theoretical yield is the predicted amount of product made if the reaction happens with a 100% yield. To find the yield, we need to know how much reactant used and then multiply it with the stoichiometric coefficient of the product. If Fe(OH)2 the limiting reactant then it means all 0.03555 mol of it will be used. The amount of iron (iii) phosphate made will be:
0.03555 mol * 1/3= 0.01185 moles
