In classic schemes the precipitate of MgNH_4PO_4 middot 6H_2O would he converted

Question

In classic schemes the precipitate of MgNH_4PO_4 middot 6H_2O would he converted by ignition to magnesium pyrophosphate, Mg_2P_2O_7, and weighed. NH_3 and H_2O were also produced. Fortunately, the MgNH_4PO_4 middot 6H_2O is sufficiently stable at room temperature to he dried and weighed, thus avoiding the ignition and the difficulties associated with it. a. Complete and balance the equation for this reaction. 2MgNH_4PO_4 middot 6H_2O(s) If 5.00 g of MgNH_4PO_4 middot 6H_2O are ignited, how many grams of Mg_2P_2O_7 would be formed? Show you calculations.

Explanation / Answer

Balance:

H2O will become vapo

so

2MgNH4PO4*6H2O = MgP2O7 (s) * 12H2O(g)

2 MgNH4PO4*6H2O = 2NH3 + 13H2O + Mg2P2O7

b)

mol of MgNH4PO4*6H2O = 137.3148+6*18 = 245.3148 g/mol

so

mol = mass/MW = 5/245.3148 = 0.02038

so

ratio is 2:1

0.0203 mol of smaple wil give --> 1/2*0.0203 = 0.01015 mol of Mg2P2O7

Mass = mol*MW = 0.01015*222.5533 = 2.2589 g will be produced

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