Question: A certain sample of phosphate rock contains 26.26% P2O5. A .5428 gram sample is analyzed by percipitating MgNH4PO4*6H2O and heating it strongly to form Mg2P2O7. compute the mass of Mg2P2O7.
I'm not entirly sure what to do. Do you set up an equation?
You may want to write equations but it isn't necessary.
I would convert 0.5428 g sample to g P2O5.
0.5428 x 0.2626 = ??
Then ?? x (1 mol P2O5/molar mass P2O5) converts that to say x mols P2O5.
If we then recognize that 1 mole P2O5 must be present to make 1 mole Mg2P2O7, then x mol P2O5 x (1 mol Mg2P2O7/1 mole P2O5) x (molar mass Mg2P2O7/1 mol Mg2P2O7) should give grams Mg2P2O7. Check my work. There is an easier way to do this but it isn't taught anymore. The OLD way (when I was a student) was to use a gravimetric factor. That converts directly as
0.5428 x 0.2626 x (molar mass Mg2P2O7/molar mass P2O5) = ??
I get something like 0.22 something.
Well, analyzing phosphate rock sounds like a rocky situation. But don't worry, I'm here to make it less intimidating for you!
To tackle this problem, let's break it down step by step:
Step 1: Calculate the amount of P2O5 in the .5428 gram sample.
The percentage of P2O5 in the rock is given as 26.26%. So, we can calculate the mass of P2O5 present in the .5428 gram sample by multiplying the percentage by the mass of the sample:
P2O5 mass = 26.26/100 * 0.5428 grams
Step 2: Convert the mass of P2O5 to the mass of Mg2P2O7.
Now, to find the mass of Mg2P2O7, we need to use the stoichiometry of the reaction. From the balanced equation, we know that 1 mole of P2O5 reacts to form 1 mole of Mg2P2O7. So, the molar mass of P2O5 will be equal to the molar mass of Mg2P2O7.
Step 3: Calculate the molar mass of Mg2P2O7.
Mg: Atomic mass = 24.31 g/mol
P: Atomic mass = 30.97 g/mol
O: Atomic mass = 16.00 g/mol
Now, Mg2P2O7 consists of two Mg atoms, two P atoms, and seven O atoms. So, we can calculate the molar mass of Mg2P2O7 using the atomic masses of Mg, P, and O:
Molar mass of Mg2P2O7 = 2 * (2 * Mg) + 2 * (1 * P) + 7 * (7 * O)
Step 4: Calculate the mass of Mg2P2O7.
To find the mass of Mg2P2O7, divide the mass of P2O5 (from step 1) by the molar mass of Mg2P2O7 (from step 3).
Now, all you need to do is plug in the numbers and do the calculations, and you'll have the mass of Mg2P2O7 in no time!
To solve this problem, we need to set up a stoichiometric equation using the given information.
First, let's recall that P2O5 is a compound consisting of two moles of phosphorus (P) and five moles of oxygen (O). Therefore, the molar mass of P2O5 can be calculated as follows:
Molar mass of P2O5 = (2 * molar mass of P) + (5 * molar mass of O)
Next, we can determine the number of moles of P2O5 in the .5428 gram sample:
Moles of P2O5 = mass of sample / molar mass of P2O5
Now, we need to write a stoichiometric equation to find the moles of Mg2P2O7 formed when P2O5 reacts with MgNH4PO4*6H2O. The balanced equation is as follows:
3 MgNH4PO4*6H2O + 4 P2O5 → 2 Mg2P2O7 + 21 H2O + 3 NH3 + 6 H2O
The stoichiometric ratio of P2O5 to Mg2P2O7 is 4:2, which simplifies to 2:1. Therefore, the moles of Mg2P2O7 formed can be determined as:
Moles of Mg2P2O7 = (Moles of P2O5) * (2 / 4)
Finally, we can calculate the mass of Mg2P2O7 formed using the molar mass of Mg2P2O7:
Mass of Mg2P2O7 = (Moles of Mg2P2O7) * (molar mass of Mg2P2O7)
By following these steps, you should be able to calculate the mass of Mg2P2O7. Let me know if you need assistance with any specific calculations!
To compute the mass of Mg2P2O7, we need to follow a series of steps based on the given information. Here's how we can approach the problem:
1. Begin by converting the percentage of P2O5 in the phosphate rock to grams. Assume we have 100 grams of the phosphate rock, which means it contains 26.26 grams of P2O5 (since it is 26.26% of 100 grams).
2. Next, calculate the molar mass of P2O5 by adding up the atomic masses of its atoms. Oxygen (O) has a mass of 16 grams/mol, and phosphorus (P) has a mass of 31 grams/mol. Multiply the atomic mass of phosphorus by 2 since there are two phosphorus atoms in P2O5 and then add the combined value of the atomic masses of oxygen. The molar mass of P2O5 is 2(31 g/mol) + 5(16 g/mol) = 142 g/mol.
3. Convert the grams of P2O5 to moles by dividing the mass (26.26 g) by the molar mass (142 g/mol). The result is approximately 0.185 moles.
4. Use the balanced chemical equation for the precipitation reaction to determine the stoichiometric ratio between P2O5 and Mg2P2O7. The balanced equation for this reaction is:
MgNH4PO4·6H2O + heat → Mg2P2O7 + 2NH3 + 9H2O
By examining the equation, we can see that 1 mole of P2O5 reacts to form 1 mole of Mg2P2O7.
5. Lastly, use the conversion factor derived from the stoichiometric ratio to determine the moles of Mg2P2O7. Since we know that 0.185 moles of P2O5 will react to form 0.185 moles of Mg2P2O7.
6. Finally, we need to calculate the mass of Mg2P2O7. Knowing the moles of Mg2P2O7 (0.185 moles), we can use its molar mass to convert to grams. The molar mass of Mg2P2O7 is calculated by adding up the atomic masses. Magnesium (Mg) has a mass of 24.3 grams/mol, and phosphorus (P) has a mass of 30.9 grams/mol. Multiply the atomic mass of phosphorus by 2 since there are two phosphorus atoms in Mg2P2O7 and multiply the atomic mass of magnesium by 2. Then, add up all the atomic masses. The molar mass of Mg2P2O7 is 2(24.3 g/mol) + 2(30.9 g/mol) + 7(16 g/mol) = 222.6 g/mol.
Multiply the moles of Mg2P2O7 (0.185 moles) by its molar mass (222.6 g/mol) to find the mass of Mg2P2O7. The result is approximately 41.19 grams. Therefore, the mass of Mg2P2O7 is 41.19 grams.