How many grams of oxygen can be produced from 165 grams of Na3PO4?
In what reaction?
To determine how many grams of oxygen can be produced from 165 grams of Na3PO4, we need to utilize the stoichiometry of the balanced equation for the reaction.
First, we need to write down the balanced equation for the reaction in which Na3PO4 is decomposed into its constituent elements:
2Na3PO4 → 6Na + P2O5 + 5O2
From the equation, we can see that for every 2 moles of Na3PO4, we get 5 moles of oxygen.
Now, we need to convert the given mass of Na3PO4 (165 grams) into moles. We can do this by using the molar mass of Na3PO4, which is calculated by adding up the atomic masses of Na (sodium), P (phosphorus), and O (oxygen).
The atomic masses are:
Na = 22.99 g/mol
P = 30.97 g/mol
O = 16.00 g/mol
So, the molar mass of Na3PO4 is:
3(22.99) + 30.97 + 4(16.00) = 163.94 g/mol
Now, we can calculate the number of moles of Na3PO4 by dividing the given mass by its molar mass:
165 g / 163.94 g/mol = 1.005 mol
Using the stoichiometry from the balanced equation, we know that 2 moles of Na3PO4 yield 5 moles of oxygen. Therefore, we can set up a proportion to find the number of moles of oxygen produced from 1.005 mol of Na3PO4:
2 mol Na3PO4 / 5 mol O2 = 1.005 mol Na3PO4 / x mol O2
Cross-multiplying, we get:
2x = 5 * 1.005
2x = 5.025
x = 2.5125
Thus, 1.005 mol of Na3PO4 would produce approximately 2.5125 mol of oxygen.
Finally, to convert the moles of oxygen to grams, we use the molar mass of oxygen (O):
Molar mass of O = 16.00 g/mol
So, the mass of oxygen produced from 165 grams of Na3PO4 would be:
2.5125 mol * 16.00 g/mol = 40.20 grams
Therefore, approximately 40.20 grams of oxygen can be produced from 165 grams of Na3PO4.