Calculate the theoretical yield of the precipitate formed when a solution containing 1.02 grams of lithium phosphate is mixed with a solution containing 2.72 grams of strontium nitrate
Why did the chemist refuse to calculate the theoretical yield of the precipitate? Because he got fired for mixing up lithium phosphide with lithium phosphate! Such a shocking formula mix-up! Good thing I'm here to help.
To calculate the theoretical yield of the precipitate, we first need to know the balanced chemical equation for the reaction between lithium phosphate (Li₃PO₄) and strontium nitrate (Sr(NO₃)₂). Do you happen to have that information?
To calculate the theoretical yield of the precipitate formed, we first need to determine which compound will precipitate when the two solutions are mixed.
To do this, we can write out the balanced chemical equation for the reaction between lithium phosphate (Li3PO4) and strontium nitrate (Sr(NO3)2):
3Li3PO4 + Sr(NO3)2 -> Sr3(PO4)2 + 6LiNO3
From the balanced equation, we can see that strontium phosphate (Sr3(PO4)2) will be formed as the precipitate.
Next, we need to determine the molar masses of lithium phosphate and strontium nitrate:
- Molar mass of Li3PO4 = (6 × lithium atomic mass) + phosphorus atomic mass + (4 × oxygen atomic mass)
- Molar mass of Sr(NO3)2 = strontium atomic mass + (2 × nitrogen atomic mass) + (6 × oxygen atomic mass)
Using the periodic table, we can find the atomic masses:
- Lithium (Li) = 6.94 g/mol
- Phosphorus (P) = 30.97 g/mol
- Oxygen (O) = 16.00 g/mol
- Strontium (Sr) = 87.62 g/mol
- Nitrogen (N) = 14.01 g/mol
Now we can calculate the molar masses:
- Molar mass of Li3PO4 = (6 × 6.94) + 30.97 + (4 × 16.00) = 115.79 g/mol
- Molar mass of Sr(NO3)2 = 87.62 + (2 × 14.01) + (6 × 16.00) = 211.63 g/mol
Now, we can calculate the number of moles of each compound:
- Moles of Li3PO4 = mass / molar mass = 1.02 g / 115.79 g/mol ≈ 0.0088 mol
- Moles of Sr(NO3)2 = mass / molar mass = 2.72 g / 211.63 g/mol ≈ 0.0128 mol
From the balanced equation, we can see that 3 moles of Li3PO4 react with 1 mole of Sr(NO3)2 to form 1 mole of Sr3(PO4)2. Therefore, the molar ratio between Li3PO4 and Sr3(PO4)2 is 3:1.
Since the molar ratio is 3:1, the number of moles of Sr3(PO4)2 produced will be 0.0088 mol × (1 / 3) = 0.00293 mol.
Finally, we can calculate the theoretical yield of the precipitate in grams by multiplying the number of moles by the molar mass of Sr3(PO4)2:
Theoretical yield = moles × molar mass = 0.00293 mol × (3 × 16.00 + 2 × 30.97) g/mol ≈ 0.357 g
Therefore, the theoretical yield of the precipitate formed when these solutions are mixed is approximately 0.357 grams.