how much phosphorus can be obtained from 225g of ammonium phosphate?

Answer 1

These problems are most easily explained by using % composition

ammonium phosphate = (NH₄)₃PO₄

atomic mass P = 30.97 g/mol

molar mass (NH4)₃PO₄ = 149.1 g/mol

Since there is 1 mol P in each mol of (NH₄)₃PO₄ , the % P = 30.97 g / 149.1 g (x100%) = 20.77% P

How much P is in 225 g (NH₄)₃PO₄?

225 g x 0.2077 = 46.7 g P (to 3 sig. figs.)

Answer 2

Atomic mass of P = 30.97 g

Molecular mass of ammonium phosphate = 149.09 g

Now,

Amount of P obtained from 149.09 g of ammonium phosphate = 30.97 g

Amount of P obtained from 225 g of ammonium phosphate = (30.97/149.09) * 225

= 46.738 g of P