2KClO3 + heat -> 2KCl + 3O2
14 moles KClO3 (3 mole O2/2 mole KClO3)
= 21 moles oxygen made
This is a common industrial method of producing oxygen.
2KClO3 --> 2KCl + 3O2For every 3 moles of oxygen gas produced, 2 moles of potassium chlorate are used.6 moles O2 * (2 moles KClO3 reacted / 3 moles O2 produced) = 4 moles KClO3
2 to 3, because of the balanced equation:2 KClO3 --> 2 KCl + 3 O2
In 2 moles of potassium dichromate, there are 16 moles of oxygen atoms (from the two oxygen atoms in each formula unit). The molar mass of oxygen is 16 g/mol, so in 2 moles of potassium dichromate, there are 32 grams of oxygen.
The balanced equation for this reaction is: 2K3PO4 + 3Al(NO3)3 -> 6KNO3 + AlPO4. This indicates that 2 moles of potassium phosphate react with 2 moles of aluminum nitrate to produce 6 moles of potassium nitrate.
The molar volume of gas at STP is 22.4 L/mol. So, 200 L of oxygen is equivalent to 200/22.4 = 8.93 moles. The balanced chemical equation for the decomposition of ammonium chlorate is 2NH4ClO3 -> 2N2 + Cl2 + 3O2. Therefore, to produce 3 moles of oxygen, we need 2 moles of ammonium chlorate. 8.93 moles of oxygen would require 2/3 * 8.93 = 5.95 moles of ammonium chlorate. The molar mass of NH4ClO3 is 101.5 g/mol, so the mass of ammonium chlorate needed would be 5.95 * 101.5 = 604.5 grams.
12 moles KClO3 (3 moles O/1 mole KClO3) = 36 moles of oxygen.
The chemical reactin is:2 KClO3 = 2 KCl + 3 O24 moles of potassium chlorate produce 6 moles oxygen.
For every mole of potassium chlorate that decomposes, three moles of oxygen are produced. Therefore, if 7.5 moles of potassium chlorate decompose, 22.5 moles of oxygen would be produced (7.5 moles x 3).
The answer is 1 mole potassium chlorate.
Four moles of potassium chlorate are needed.
For the decomposition of potassium chlorate, the molar ratio between potassium chlorate (KClO3) and oxygen (O2) is 2:3. Therefore, to produce 15 moles of oxygen, 10 moles of potassium chlorate are needed. (15 moles O2) x (2 moles KClO3 / 3 moles O2) = 10 moles KClO3.
To determine the grams of potassium chloride formed, you first need to calculate the moles of oxygen produced by the decomposition of potassium chlorate. Then, use the stoichiometry of the balanced chemical equation to convert moles of oxygen to moles of potassium chloride. Finally, from the molar mass of potassium chloride, you can calculate the grams formed.
1 mole of potassium chlorate produces 3 moles of oxygen gas when heated, or 1 mole of potassium chlorate produces 1.344 L of oxygen gas at NTP. To produce 2.24 L of oxygen gas, you would need about 1.67 moles of potassium chlorate.
To calculate the amount of oxygen gas produced from potassium chlorate, use the balanced chemical equation for the decomposition of potassium chlorate: 2KClO3 -> 2KCl + 3O2. From the equation, every 2 moles of KClO3 produce 3 moles of O2. First, convert the given mass of KClO3 to moles, then use the mole ratio from the balanced equation to find the moles of O2 produced. Finally, convert moles of O2 to grams using its molar mass (32 g/mol).
2KClO3 --> 2KCl + 3O2For every 3 moles of oxygen gas produced, 2 moles of potassium chlorate are used.6 moles O2 * (2 moles KClO3 reacted / 3 moles O2 produced) = 4 moles KClO3
We need 3 moles of potassium perchlorate.
The balanced chemical equation for the decomposition of potassium chlorate (KClO3) into oxygen (O2) and potassium chloride (KCl) is 2KClO3 -> 2KCl + 3O2. This means that for every 2 moles of KClO3, 2 moles of KCl and 3 moles of O2 are produced.