Mass of small beaker 112.64 g Mass of Fertilizer and Beaker 123.03 g Mass of fer
ID: 548444 • Letter: M
Question
Mass of small beaker
112.64 g
Mass of Fertilizer and Beaker
123.03 g
Mass of fertilizer sample
10.39 g
Mass of filter paper
0.44 g
Mass of filter paper and dry MgNH4PO4*6H2O precipitate
0.96 g
Mass of dry MgNH4PO4*6H2O precipitate
0.52 g
Moles of MgNH4PO4*6H2O
0.0021 mol
Moles of Phosphorous
0.0021 mol
Mass of Phosphorous
0.0065 g
% P in fertilizer sample
.62 %
Moles of P2O5
mol
Mass of P2O5
g
% as P2O5 in fertilizer sample
%
Mass of small beaker
112.64 g
Mass of Fertilizer and Beaker
123.03 g
Mass of fertilizer sample
10.39 g
Mass of filter paper
0.44 g
Mass of filter paper and dry MgNH4PO4*6H2O precipitate
0.96 g
Mass of dry MgNH4PO4*6H2O precipitate
0.52 g
Moles of MgNH4PO4*6H2O
0.0021 mol
Moles of Phosphorous
0.0021 mol
Mass of Phosphorous
0.0065 g
% P in fertilizer sample
.62 %
Moles of P2O5
mol
Mass of P2O5
g
% as P2O5 in fertilizer sample
%
Explanation / Answer
It is assumed that P2O5 is the source of P in the fertilizer. Assume a hypothetical decomposition of P2O5 (without charges on the ions).
P2O5 -------> 2 P + 5 O
As per the stoichiometric equation above,
1 mole P2O5 = 2 moles P.
We have deduced earlier that the fertilizer contained 0.0021 mole P; therefore, mole(s) of P2O5 = (0.0021 mole P)*(1 mole P2O5/2 mole P) = 0.00105 mole P2O5 (ans)
Molar mass of P2O5 = (2*30.9738 + 5*15.9994) g/mol = 141.9446 g/mol.
Mass of P2O5 in the fertilizer = (0.00105 mole)*(141.9446 g/mol) = 0.14904 g 0.149 g (ans).
% P2O5 in the fertilizer = (0.149 g)/(10.39 g)*100 = 1.434% (ans).