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How many moles of oxygen are produced from twelve moles of potassium chlorate?
12 moles KClO3 (3 moles O/1 mole KClO3) = 36 moles of oxygen.
How many moles of potassium chlorate are needed to produce 15 moles of oxygen?
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.
How much potassium chlorate should be heated to produce 2.24 L of oxygen at NTP?
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.
How many moles of oxygen are produced when 7.5 moles of potassium chlorate decompose completely?
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).
How much oxygen gas in grams can be prepared from 86.8 g 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).
How many moles of potassium chlorate must be used to produce 6 moles if oxygen gas?
Four moles of potassium chlorate are needed.
How many moles of oxygen are produced from twelve moles of potassium chlorate?
12 moles KClO3 (3 moles O/1 mole KClO3) = 36 moles of oxygen.
How many moles of potassium chlorate are needed to produce 15 moles of oxygen?
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.
How much potassium chlorate should be heated to produce 2.24 L of oxygen at NTP?
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.
How many moles potassium chlorate must be used to produce 6 moles of oxygen gas?
We need 3 moles of potassium perchlorate.
How many moles of potassium chlorate are needed to give 1.50 moles of oxygen?
The answer is 1 mole potassium chlorate.
How many moles of oxygen are produced when 7.5 moles of potassium chlorate decompose completely?
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).
How many moles of potassium chlorate (KClo3) must be used to produce 6 moles of oxygen?
2KClO3==>2KCl+3O2 is the equation. so you need 4 moles of KClO3.
How many moles of oxygen are produced from 14 moles of potassium chlorate?
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.
How many moles of potassium chlorate must be used to produce 6 moles of oxygen gas?
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
How much oxygen gas in grams can be prepared from 86.8 g 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).
How many moles of cesium chlorate must be decomposed to produce 2.7 moles of oxygen gas?
To find the moles of cesium chlorate needed to produce 2.7 moles of oxygen gas, use the balanced chemical equation for the decomposition of cesium chlorate: 2CsClO3 -> 2CsCl + 3O2 From the equation, it shows that 2 moles of cesium chlorate produce 3 moles of oxygen gas. Therefore, you will need (2/3) * 2.7 = 1.8 moles of cesium chlorate to produce 2.7 moles of oxygen gas.
