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19 October, 04:36

A student uses 3.00g of a fertilizer with an npk ratio of 15-15-10, how much mgnh4po4*6h2o (mm = 245.1 g/mol) should the student produced?

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  1. 19 October, 04:42
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    Supposing all of the P in the fertilizer is converted PO4 anions in the Mg salt:

    (3.00 g) x (0.15) / (30.97376 g P/mol) x (1 mol MgNH4PO4*6H2O / 1 mol P) x (245.1 g MgNH4PO4*6H2O/mol) =

    3.56 g MgNH4PO4*6H2O
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