Lithium chlorate is decomposed with heat to give lithium chloride and oxygen gas. If 2.23g of lithium chlorate is decomposed, how many milliliters of oxygen are released at stp?
write the balanced equation
2LiCl03>2LiCl + 3O2
2.23 g of liCl03:figure moles of LiCl03
so you will get 1.5 times those moles of O2. Multply by 22.4 to get liters, then a 1000 to get milliliters.
To find the number of milliliters of oxygen gas released during the decomposition of lithium chlorate, we need to use stoichiometry and the ideal gas law.
1. Write the balanced chemical equation for the decomposition of lithium chlorate:
2LiClO3(s) → 2LiCl(s) + 3O2(g)
2. Calculate the number of moles of lithium chlorate decomposed:
Given mass of lithium chlorate = 2.23g
Molar mass of lithium chlorate (LiClO3) = 6.941 + 35.453 + 48 = 90.394 g/mol
Number of moles of lithium chlorate = mass / molar mass = 2.23g / 90.394 g/mol
3. Determine the number of moles of oxygen gas produced:
From the balanced equation, we see that 2 moles of lithium chlorate produce 3 moles of oxygen gas.
So, moles of oxygen gas = 3/2 * moles of lithium chlorate
4. Convert the number of moles of oxygen gas to the volume at STP:
1 mole of any ideal gas at STP occupies 22.4 L (or 22,400 mL).
Volume of oxygen gas at STP = moles of oxygen gas * 22,400 mL/mol
5. Plug in the known values and calculate the volume of oxygen gas released:
Volume of oxygen gas = (3/2 * moles of lithium chlorate) * 22,400 mL/mol
Now you can substitute the value of moles of lithium chlorate and calculate the volume of oxygen gas released.
To find the number of milliliters of oxygen gas released, we need to use the ideal gas law to calculate the volume of gas at STP (Standard Temperature and Pressure).
1. Start by calculating the number of moles of lithium chlorate (LiClO3):
- The molar mass of LiClO3 is: 6.94 g/mol (lithium) + 35.45 g/mol (chlorine) + 3 * 16.00 g/mol (oxygen) = 106.45 g/mol.
- Divide the given mass (2.23 g) by the molar mass to find the number of moles: 2.23 g / 106.45 g/mol ≈ 0.021 moles.
2. According to the balanced chemical equation, 1 mole of LiClO3 decomposes to produce 1 mole of O2 gas:
- This means that the number of moles of O2 gas released is also 0.021 moles.
3. Now, we can use the ideal gas law, PV = nRT, to calculate the volume of the gas at STP. At STP, the temperature (T) is 273.15 K, and the pressure (P) is 1 atm.
- Rearrange the equation to solve for V: V = (nRT) / P.
- Plug in the values: V = (0.021 moles * 0.0821 L·atm/mol·K * 273.15 K) / 1 atm.
- Calculate: V ≈ 0.478 L.
4. Convert the volume from liters to milliliters:
- Since 1 L = 1000 mL, multiply the volume by 1000: 0.478 L * 1000 mL/L ≈ 478 mL.
Therefore, approximately 478 milliliters of oxygen gas would be released at STP when 2.23 grams of lithium chlorate is decomposed.