For this problem, will someone please check my answer?
How many grams of oxygen are produced in the decomposition of 5.00g of potassium chlorate?
KClO3(s)--->KCl(s)+O2(g)
[the equation may not be balanced]
My answer is 1.96 g of O2.
The balanced equation is
2KClO3 ==> 2KCl + 3O2
I, too, obtained 1.96g O2. You must have balanced the equation before solving, else our answers would not have agreed.
Yes, I did. Thank you!
To find the amount of grams of oxygen produced in the decomposition of potassium chlorate, you first need to balance the equation. The balanced equation is:
2KClO3 → 2KCl + 3O2
Now that the equation is balanced, you can use stoichiometry to calculate the grams of oxygen produced.
Given that you have 5.00g of potassium chlorate (KClO3), you can convert this mass to moles using the molar mass of KClO3. The molar mass of KClO3 is:
K (39.10 g/mol) + Cl (35.45 g/mol) + 3O (16.00 g/mol) = 122.55 g/mol
5.00g KClO3 * (1 mol KClO3 / 122.55 g KClO3) = 0.0408 mol KClO3
According to the balanced equation, you know that for every 2 moles of KClO3, you produce 3 moles of O2. Therefore, you can set up a ratio:
0.0408 mol KClO3 * (3 mol O2 / 2 mol KClO3) = 0.0612 mol O2
Now, to calculate the grams of oxygen, you can multiply the moles of O2 by the molar mass of O2:
0.0612 mol O2 * (32.00 g/mol O2) = 1.96 g O2
So, your answer of 1.96 g of O2 is correct. Well done!