For the following reaction, 24

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