Chemistry work........the volume of oxygen measured at stp will br produced on heating 24.5grams of potassium trioxochlorate(v)
Your School SUBJECT seems to be CHEMISTRY.
2KClO3 ==> 2KCl + 3O2
mols KClO3 = grams/molar mass = ?
Using the coefficients in the balanced equation, convert mols KClO3 to mols O2.
Then mols O2 x 22.4L = L O2 at STP
Kclo3----------- 2lcl+302
To determine the volume of oxygen produced when heating potassium trioxochlorate(V), you need to follow these steps:
Step 1: Determine the molar mass of potassium trioxochlorate(V).
Potassium trioxochlorate(V) has the formula KClO3. The molar mass of potassium (K) is 39.10 g/mol, and the molar mass of chlorine (Cl) is 35.45 g/mol. The molar mass of oxygen (O) is 16.00 g/mol. Add up the molar masses:
(39.10 g/mol potassium) + (1 * 35.45 g/mol chlorine) + (3 * 16.00 g/mol oxygen) = 39.10 + 35.45 + 48.00 = 122.55 g/mol.
Step 2: Calculate the number of moles of potassium trioxochlorate(V).
To convert grams to moles, divide the mass of the substance by its molar mass:
Number of moles = Mass (g) / Molar mass (g/mol)
Number of moles = 24.5 g / 122.55 g/mol = 0.20 mol (rounded to two significant figures).
Step 3: Use the balanced chemical equation.
The balanced chemical equation for the thermal decomposition of potassium trioxochlorate(V) is 2KClO3 → 2KCl + 3O2.
From the balanced equation, you can see that for every 2 moles of KClO3, 3 moles of O2 are produced.
Step 4: Calculate the molar ratio.
Since we know that 0.20 moles of KClO3 are used, we can set up a proportion to find the number of moles of O2:
(0.20 mol KClO3 / 2 mol KClO3) = (x mol O2 / 3 mol O2)
Solving for x gives us:
x = (0.20 mol KClO3 * 3 mol O2) / 2 mol KClO3 = 0.30 mol O2.
Step 5: Use the ideal gas law.
At STP (Standard Temperature and Pressure), 1 mole of any ideal gas occupies 22.4 liters. Therefore, 0.30 moles of O2 would occupy:
Volume (L) = 0.30 mol O2 * 22.4 L/mol = 6.72 L (rounded to two significant figures).
So, when heating 24.5 grams of potassium trioxochlorate(V), the volume of oxygen produced at STP is approximately 6.72 liters.