Calculate how many grams of O2(g) can be produced from heating 65.8 grams of KClO3(s).

Well, let's break it down. The molar mass of KClO3 is 122.55 g/mol. So, in 65.8 grams of KClO3, we have 65.8 g / 122.55 g/mol = 0.5372 mol of KClO3.

The balanced equation for the reaction of KClO3 decomposing into O2 is 2 KClO3 -> 2 KCl + 3 O2. So, for every 2 moles of KClO3, we get 3 moles of O2.

Therefore, we can calculate the moles of O2 produced from 0.5372 mol of KClO3: 0.5372 mol KClO3 * (3 mol O2 / 2 mol KClO3) = 0.8058 mol O2.

Now, to calculate the mass of O2, we need to know its molar mass, which is approximately 32 g/mol. So, 0.8058 mol O2 * 32 g/mol = 25.78 grams of O2.

So, heating 65.8 grams of KClO3 would produce approximately 25.78 grams of O2. That's enough oxygen to inflate quite a few balloons or maybe help you blow out some candles on a cake!

To calculate the number of grams of O2(g) that can be produced from heating KClO3(s), we need to use the balanced chemical equation for the reaction.

The balanced chemical equation is:

2 KClO3(s) → 2 KCl(s) + 3 O2(g)

From the equation, we can see that 2 moles of KClO3 will produce 3 moles of O2.

To find the number of moles of KClO3, we use its molar mass. The molar mass of KClO3 is calculated as follows:
K: 39.10 g/mol
Cl: 35.45 g/mol
O: 16.00 g/mol

Molar mass of KClO3 = (39.10 g/mol) + (35.45 g/mol) + (16.00 g/mol x 3)
= 39.10 g/mol + 35.45 g/mol + 48.00 g/mol
= 122.55 g/mol

Next, we calculate the number of moles of KClO3 by dividing the given mass by the molar mass:
Number of moles = mass / molar mass
= 65.8 g / 122.55 g/mol
= 0.537 mol

According to the stoichiometry of the balanced equation, 2 moles of KClO3 will produce 3 moles of O2.

Using the mole ratios, we can calculate the number of moles of O2 produced:
Number of moles of O2 = 0.537 mol x (3 mol O2 / 2 mol KClO3)
= 0.8055 mol

Finally, we can calculate the mass of O2 produced using its molar mass:
Mass of O2 = number of moles x molar mass
= 0.8055 mol x 32.00 g/mol
= 25.78 g

Therefore, heating 65.8 grams of KClO3 will produce approximately 25.78 grams of O2.

To calculate the number of grams of O2(g) produced from heating a given mass of KClO3(s), we need to use stoichiometry. The balanced chemical equation for the reaction is:

2 KClO3(s) -> 2 KCl(s) + 3 O2(g)

From the balanced equation, we can see that for every 2 moles of KClO3, 3 moles of O2 are produced.

To find the number of moles of KClO3, we can use the molar mass of KClO3:

Molar mass of KClO3 = Molar mass of K (39.10 g/mol) + Molar mass of Cl (35.45 g/mol) + 3 * Molar mass of O (16.00 g/mol)
= 39.10 g/mol + 35.45 g/mol + 48.00 g/mol
= 122.55 g/mol

Now, we can calculate the number of moles of KClO3:

Moles of KClO3 = Mass of KClO3 / Molar mass of KClO3
= 65.8 g / 122.55 g/mol
= 0.537 mol

According to the stoichiometry of the balanced equation, for every 2 moles of KClO3, 3 moles of O2 are produced. Therefore, we can find the number of moles of O2 produced:

Moles of O2 = Moles of KClO3 * (3 moles O2 / 2 moles KClO3)
= 0.537 mol * (3 mol O2 / 2 mol KClO3)
= 0.8055 mol O2

Finally, we can calculate the mass of O2 in grams:

Mass of O2 = Moles of O2 * Molar mass of O2
= 0.8055 mol * 32.00 g/mol
= 25.77 g

Therefore, heating 65.8 grams of KClO3 will produce approximately 25.77 grams of O2.

2KClO3>>3O2+2KCl

so for every mole of KClO3, you get 3/2 mole O2

figure moles of KClO3, multiply by 3/2, then multiply by 32 to get grams oxgen gas.