Calculate how many grams of oxygen form when 2.72 g of KCIOg reacts completely given the
1 answer:
1.06g
Please refer to the image attached.
You first find the number of moles (n) for KClO3 as you have been given the mass. So n = mass/Mr gives you the moles that is 0.022 mol
Next you use the ratio 2:3 as per the equation where it's 2KClO3 ---> 3O2 so you just cross multiply and get the moles for oxygen.
Now that you know the number of moles for oxygen, you can use the same moles equation that you used earlier to find the mass. mass=n×Mr and so you get the answer as 1.06g of oxygen.
You might be interested in
Answer:
<u></u>
- <u>0.456M</u>
Explanation:
<u>1. Balanced molecular equation</u>
<u>2. Mole ratio</u>
<u>3. Moles of HNO₃</u>
- Number of moles = Molarity × Volume in liters
- n = 0.600M × 0.0100 liter = 0.00600 mol HNO₃
<u>4. Moles Ba(OH)₂</u>
- n = 0.700M × 0.0310 liter = 0.0217 mol
<u>5. Limiting reactant</u>
Actual ratio:
Since the ratio of the moles of HNO₃ available to the moles of Ba(OH)₂ available is less than the theoretical mole ratio, HNO₃ is the limiting reactant.
Thus, 0.006 moles of HNO₃ will react completely with 0.003 moles of Ba(OH)₂ and 0.0217 - 0.003 = 0.0187 moles will be left over.
<u>6. Final molarity of Ba(OH)₂</u>
- Molarity = number of moles / volume in liters
- Molarity = 0.0187 mol / (0.0100 + 0.0031) liter = 0.456M