Question

If you collected a 3.02g sample of MgNH4PO4x6H2O calculate a the value for the mass of P2O5 in the initial mass of fertilizer analyzed. What is the % P2O5 in the fertilizer? a) 0.747 g P2O5; 24.71%

b) 0.451 g P2O5; 15.78%
c) 0.602 g P2O5; 19.93%
d) 1.204 g P2O5; 39.86%

Answer 1

To calculate the mass of P2O5 in the fertilizer, we can use stoichiometry and the molar mass of MgNH4PO4·6H2O. The mass of P2O5 is 0.76 g and the % P2O5 in the fertilizer is 25.17%.

Step-by-step explanation:

To calculate the mass of P2O5 in the initial mass of the fertilizer, we need to determine the molar mass of MgNH4PO4·6H2O and then use stoichiometry. The molar mass of MgNH4PO4·6H2O is:

Mg: 24.31 g/mol

N: 14.01 g/mol

H: 1.008 g/mol

P: 30.97 g/mol

O: 16.00 g/mol

The total molar mass is then:

Total mass = (1 × 24.31) + (4 × 14.01) + (12 × 1.008) + (1 × 30.97) + (6 × 16.00)

= 246.10 g/mol

Next, we can use stoichiometry to convert the mass of MgNH4PO4·6H2O to the mass of P2O5. The balanced equation for the reaction is:

(NH4)3PO4 + MgSO4 → MgNH4PO4 + (NH4)2SO4

From the equation, we can see that 1 mol of MgNH4PO4 corresponds to 2 mol of P2O5. Therefore, the molar mass of P2O5 is:

Molar mass of P2O5 = (2 × 30.97)

= 61.94 g/mol

Finally, we can calculate the mass of P2O5:

Mass of P2O5 = (Mass of MgNH4PO4) × (Molar mass of P2O5 / Molar mass of MgNH4PO4)

= (3.02 g) × (61.94 g/mol / 246.10 g/mol)

= 0.76 g

The % P2O5 in the fertilizer is:

% P2O5 = (Mass of P2O5 / Mass of fertilizer) × 100%

= (0.76 g / 3.02 g) × 100%

= 25.17%