Final answer:
Upon heating MgNH4PO4 · 6H2O, ammonia, water, and magnesium pyrophosphate (Mg2P2O7) are formed. By using the molar masses of the substances and the balanced equation, we can calculate that 2.27 grams of Mg2P2O7 are produced from 2.50 grams of MgNH4PO4 · 6H2O.
Step-by-step explanation:
The ignition of MgNH4PO4 · 6H2O leads to the production of NH3, H2O, and magnesium pyrophosphate (Mg2P2O7). Upon heating, the compound decomposes, releasing ammonia and water while forming magnesium pyrophosphate. The balanced equation for this reaction is:
2 MgNH4PO4 · 6H2O(s) → 2 Mg2P2O7(s) + 4 NH3(g) + 12 H2O(g)
To calculate the mass of Mg2P2O7 formed when 2.50 g of MgNH4PO4 · 6H2O are ignited, we need the molar masses of both substances. For MgNH4PO4 · 6H2O, it is approximately 245.41 g/mol, and for Mg2P2O7, it is approximately 222.56 g/mol.
First, calculate the moles of MgNH4PO4 · 6H2O:
Moles of MgNH4PO4 · 6H2O = 2.50 g / 245.41 g/mol
Moles of MgNH4PO4 · 6H2O = 0.0102 mol
Since the mole ratio between MgNH4PO4 · 6H2O and Mg2P2O7 in the balanced equation is 1:1, the moles of Mg2P2O7 formed will also be 0.0102 mol.
Now, convert moles of Mg2P2O7 to grams:
Mass of Mg2P2O7 = 0.0102 mol * 222.56 g/mol
Mass of Mg2P2O7 = 2.27 g
Therefore, 2.27 grams of Mg2P2O7 are formed when 2.50 g of MgNH4PO4 · 6H2O are ignited.