A 2.45 gram sample of potassium chlorate is heated an the oxygen gas collected. what is the volume of the oxygen at STP.
Is the oxygen gas itself 2.45 grams or do I have to subtract something from that?
No, the KClO3 from which the oxygen is evolved, has a mass of 2.45 g. You need to write the equation, balance it, convert 2.45 g KClO3 to mols, convert mols KClO3 to mols O2 using the coefficients in the balanced equation, then convert mols O2 to liters knowing that 22.4 L is the volume occupied by a mole of a gas at STP. USUALLY, oxygen is collected over water and that must be corrected for the vapor pressure of water; however, your problem doesn't say anything about water; therefore, I asume we don't need to worry about that.Post your work if you get stuck and we can help you through it.
what reactants would there be besides O2?
Heating KClO3 produces oxygen and potassium chloride.
KClO3 + heat ==> KCl + O2
The balanced equation is:
2KClO3 ==> 2KCl + 3O2.
To find the volume of oxygen gas at STP (Standard Temperature and Pressure), we need to consider the molar mass of potassium chlorate (KClO3) and use the mole-to-volume relationship.
First, let's calculate the number of moles of potassium chlorate. We can use its molar mass, which is the sum of the atomic masses of its constituent elements: K (potassium) has a molar mass of 39.10 g/mol, Cl (chlorine) has a molar mass of 35.45 g/mol, and O (oxygen) has a molar mass of 16.00 g/mol.
Molar mass of KClO3 = 39.10 g/mol + 35.45 g/mol + (16.00 g/mol × 3) = 122.55 g/mol
Next, we can calculate the number of moles of potassium chlorate using its mass:
Number of moles = Mass / Molar mass
Number of moles = 2.45 g / 122.55 g/mol
Now that we have the number of moles of potassium chlorate, we can determine the moles of oxygen gas produced during the reaction. For every 2 moles of KClO3, we get 3 moles of O2.
So, moles of O2 = (Number of moles of KClO3) × (3 moles O2 / 2 moles KClO3)
Now, let's convert the moles of O2 to the volume at STP using the ideal gas law, which states that at STP, 1 mole of gas occupies 22.4 liters:
Volume of O2 = (Moles of O2) × (22.4 L/mol)
By following these steps, you can calculate the volume of oxygen gas at STP. Let me know if you would like me to perform the calculations for you.