In: Chemistry
For the following reaction, 24.8 grams of diphosphorus pentoxide are allowed to react with 14.0 grams of water.
diphosphorus pentoxide (s) + water (l) phosphoric acid (aq)
What is the maximum amount of phosphoric acid that can be formed?
What is the FORMULA for the limiting reagent?
What amount of the excess reagent remains after the reaction is complete?
1)
Molar mass of P2O5 = 2*MM(P) + 5*MM(O)
= 2*30.97 + 5*16.0
= 141.94 g/mol
mass of P2O5 = 24.8 g
we have below equation to be used:
number of mol of P2O5,
n = mass of P2O5/molar mass of P2O5
=(24.8 g)/(141.94 g/mol)
= 0.1747 mol
Molar mass of H2O = 2*MM(H) + 1*MM(O)
= 2*1.008 + 1*16.0
= 18.016 g/mol
mass of H2O = 14.0 g
we have below equation to be used:
number of mol of H2O,
n = mass of H2O/molar mass of H2O
=(14.0 g)/(18.016 g/mol)
= 0.7771 mol
we have the Balanced chemical equation as:
P2O5 + 3 H2O ---> 2 H3PO4
1 mol of P2O5 reacts with 3 mol of H2O
for 0.1747 mol of P2O5, 0.5242 mol of H2O is required
But we have 0.7771 mol of H2O
so, P2O5 is limiting reagent
we will use P2O5 in further calculation
Molar mass of H3PO4 = 3*MM(H) + 1*MM(P) + 4*MM(O)
= 3*1.008 + 1*30.97 + 4*16.0
= 97.994 g/mol
From balanced chemical reaction, we see that
when 1 mol of P2O5 reacts, 2 mol of H3PO4 is formed
mol of H3PO4 formed = (2/1)* moles of P2O5
= (2/1)*0.1747
= 0.3494 mol
we have below equation to be used:
mass of H3PO4 = number of mol * molar mass
= 0.3494*97.99
= 34.2 g
Answer: 34.2 g
2)
P2O5 is limiting reagent
3)
From balanced chemical reaction, we see that
when 1 mol of P2O5 reacts, 3 mol of H2O reacts
mol of H2O reacted = (3/1)* moles of P2O5
= (3/1)*0.1747
= 0.5242 mol
mol of H2O remaining = mol initially present - mol reacted
mol of H2O remaining = 0.7771 - 0.5242
mol of H2O remaining = 0.2529 mol
Molar mass of H2O = 2*MM(H) + 1*MM(O)
= 2*1.008 + 1*16.0
= 18.016 g/mol
we have below equation to be used:
mass of H2O,
m = number of mol * molar mass
= 0.2529 mol * 18.016 g/mol
= 4.56 g
Answer:4.56 g