8,55 moles of oxygen are produced.
Potassium chlorate = KClO3Decompostion: 2KClO3 ==> 3O2 + 2KCl7.5 moles KClO3 x 3 mole O2/2 moles KClO3 = 11.25 moles O2
Nothing is produced, 500g potassium chlorate will be the same 500 g potassium chlorate after reaction. Actually there is no reaction at all.
2 KClO3 ----> 2KCl + 3O2 So 2 moles of Potassium Chlorate produces 3 moles of oxygen molecules or 6 moles of oxygen atoms. 3 moles of Potassium chlorate would thus produce 4.5 moles of oxygen molecules or 9 moles of oxygen atoms.
15 moles of O2 are produced by the decomposition if 10 moles potassium chlorate 2 KClO3 and 2 KCl plus 3O2.
The anwer is 0,061 moles of oxygen.
The reaction is:2 KClO3 = 2 KCl + 3 O238,4 moles of oxygen are produced.
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
Oxygen gas is produced. The hydrogen peroxide will decompose to give water and oxygen, potassium Iodide is acting as a catalyst 2H2O2(l) ------> 2H2O(l) + O2(g)
A balanced equation for the reaction is 2 KClO3 -> 2 KCl + 3 O2. Therefore, each mole of potassium chlorate produces 3/2 moles of oxygen, and 1.2 moles of potassium chlorate produces (3/2)1.2 or 1.8 moles of oxygen.
12 moles KClO3 (3 moles O/1 mole KClO3) = 36 moles of oxygen.
The equation that describes this process is as follows: 2KClO3 ---> 2KCl + 3O2 For every 2 moles of reactants consumed 3 moles of oxygen gas are produced. 3 mol O2 / 2 mol KClO3 = x mol O2 / 12.3 mol KClO3 x = 12.3 mol x 3 mol / 2 mol = 18.45 mol Therefore, 18.5 mol (3 significant figures) of oxygen are produced by the decomposition of 12.3 mol of potassium chlorate
The reaction is:KOH + HCl = KCl + H2OThe answer is 5 moles KCl.
You have2KClO3 ==> 2KCl + 3O2 as the balanced equation 25 g KClO3 x 1 mole/123 g = 0.20 moles moles KCl formed = 0.20 moles KClO3 x 2 moles KCl/2 moles KClO3 = 0.20 moles KCl formed grams KCl = 0.20 moles x 74.5 g/mole = 14.9 g = 15 grams of KCl formed
It is a vigorous and exothermic reaction, in which a salt of Potassium and Hydrogen is produced. The heat produced may be enough to burn the hydrogen produced.
Oxygen can be prepared in the laboratory by red mercuric oxide, or by heating Potassium Chlorate in a test tube in the presence of Maganesedioxide. Fit a test tube with a delivery tube and place the delivery tube under a gas jar filled with water and placed on beehive shelve in a water tub. Heat the test tube containing Potassium Chlorate and Maganesedioxide. The oxygen produced in this reaction is passing through the delivery tube to the gas jar by the downward displacement of water.
there are two moles produced in potassium nitrate.
Equation: 2KClO3 + Cl2 ---> 2KCl + 3O2 + Cl2 1. Solve for the number of moles of KClO3 in 36.3 g. (.2962 molKClO3) 2. Multiply that value by (3/2), from the equation's coefficients. (.4447 molO2) Note: A BCA table could also be used. 3. Solve for the mass of .4447 molO2. 14.2 grams of oxygen would be produced.
A 1.80-gram mixture of potassium chlorate, kclo3, and potassium chloride, kcl, was heated until all of the kclo3 had decomposed the liberated oxygen, after drying, occupied 405 ml at 25C when the barometric pressure was 745 torr. This is the problem and the questions were... a. How many moles of O2 were produced? b. What percent of the mixture was KClO3? KCl? Please help!!
they are found in ores which are put in furnaces to extract the potassium and phosphorus.
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.
copper (thiocyanate)2 and potassium nitrate
Potassium produces a lilac flame
60 percent of the potash extracted in 2003 was produced as potassium chloride, with potassium sulfate and potassium magnesium sulfate--both for fertilizing certain crops and soils--representing the remainder
extraction from the ore via electrolysis :)
Bicarbonate and Potassium.