ignition of mgnh4po4*6h2o produces nh3, h2o and magnesium pyrophosphate, Mg2P2O7. complete and balance the equation for this reaction. If 5.00 g of MgNH4PO4*6H2O are ignited, how many grams of Mg2P2O7 would be formed?
2MgNH4PO4*6H2O(g)--->
2MgNH4PO4.6H2O ==> Mg2P2O7 + 7H2O + 2NH3
Convert 5.00 g NgNH4PO4 to mols. mols = grams/molar mass.
Use the coefficients to convert mols MgNH4PO4 to Mg2P2O7.
Now convert mols Mg2P2O7 to grams. g = mols x molar mass.
2NH3(g) + 7H2O(g) + Mg2P2O7(s)
To balance the equation, we need to determine the stoichiometric coefficients for each compound. The given equation is:
2MgNH4PO4*6H2O(g) ---> xNH3(g) + yH2O(g) + Mg2P2O7(s)
To balance the ammonium (NH4) and phosphate (PO4) ions, we need to include 2 NH3 and 1 P2O7. So, we have:
2MgNH4PO4*6H2O(g) ---> 2NH3(g) + xH2O(g) + Mg2P2O7(s)
Next, we need to balance the magnesium (Mg) and hydrogen (H) atoms. The equation now becomes:
2MgNH4PO4*6H2O(g) ---> 2NH3(g) + 12H2O(g) + Mg2P2O7(s)
To balance the oxygen (O) atoms, we calculate the number of oxygen atoms on the reactant side (left) and set it equal to the number on the product side (right). We have:
2MgNH4PO4*6H2O(g) ---> 2NH3(g) + 12H2O(g) + Mg2P2O7(s)
2 * 22 + 12 * 1 + 6 * 16 = 2 * 1 + x + 2 * 7
44 + 12 + 96 = 2 + x + 14
152 = x + 16
x = 136
The balanced equation for the reaction is:
2MgNH4PO4*6H2O(g) ---> 2NH3(g) + 12H2O(g) + 136Mg2P2O7(s)
Now, we need to calculate the molar mass of Mg2P2O7 to determine the amount formed.
Molar mass of Mg2P2O7 = (2 * Atomic mass of Mg) + (2 * Atomic mass of P) + (7 * Atomic mass of O)
= (2 * 24.31) + (2 * 30.97) + (7 * 16.00)
= 48.62 + 61.94 + 112.00
= 222.56 g/mol
To find the number of moles of Mg2P2O7 formed, we use the given mass of MgNH4PO4*6H2O and its molar mass:
Number of moles of MgNH4PO4*6H2O = Given mass / Molar mass
= 5.00 g / ((2 * 24.31) + (2 * 14.01) + (4 * 1.01) + (1 * 31.00) + (6 * 2.02) + (6 * 16.00))
= 5.00 g / (48.62 + 28.02 + 4.04 + 31.00 + 12.12 + 96.00)
= 5.00 g / 219.80
= 0.0227 mol
Since the stoichiometric coefficient of Mg2P2O7 is 136, we can calculate the number of moles of Mg2P2O7 formed:
Number of moles of Mg2P2O7 = 0.0227 mol * (136 mol Mg2P2O7 / 2 mol MgNH4PO4*6H2O)
= 1.5468 mol
Finally, to find the mass of Mg2P2O7 formed, we use the molar mass of Mg2P2O7:
Mass of Mg2P2O7 = Number of moles * Molar mass
= 1.5468 mol * 222.56 g/mol
= 343.95 g
Therefore, if 5.00 g of MgNH4PO4*6H2O are ignited, 343.95 g of Mg2P2O7 would be formed.
To balance the equation for the ignition of MgNH4PO4*6H2O, you can follow these steps:
Step 1: Break down the formula of the compound into individual ions or molecules. In this case, MgNH4PO4*6H2O can be broken down into the following components:
Mg2+
(NH4)+
PO4^3-
6H2O
Step 2: Determine the reactants and the products. The reactant in this case is MgNH4PO4*6H2O, and the products are NH3, H2O, and Mg2P2O7.
Step 3: Write the unbalanced equation by putting the reactant on the left side of the arrow and the products on the right side. The equation will look like this:
MgNH4PO4*6H2O → NH3 + H2O + Mg2P2O7
Step 4: Balance the equation by ensuring that the number of atoms of each element is the same on both sides. In this case, the balanced equation is as follows:
2MgNH4PO4*6H2O → 6NH3 + 12H2O + Mg2P2O7
Now that we have a balanced equation, we can calculate how many grams of Mg2P2O7 would be formed when 5.00 g of MgNH4PO4*6H2O is ignited. We need to use the concept of stoichiometry.
The molar mass of MgNH4PO4*6H2O is calculated as:
(2 * molar mass of Mg) + (molar mass of N) + (4 * molar mass of H) + (molar mass of P) + (4 * molar mass of O) + (6 * molar mass of H2O)
To determine the molar mass, you need to look up the individual molar masses of each element in the periodic table. Then, you can calculate it and find the number of moles in 5.00 grams.
Once you know the number of moles of MgNH4PO4*6H2O, you can use the coefficients in the balanced equation to determine the ratio between MgNH4PO4*6H2O and Mg2P2O7. In this case, the ratio is 1:1.
Finally, use the molar mass of Mg2P2O7 to calculate the number of grams formed.
I hope that helps! Let me know if you have any further questions.