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
8,55 moles of oxygen are produced.
Nothing is produced, 500g potassium chlorate will be the same 500 g potassium chlorate after reaction. Actually there is no reaction at all.
Potassium chlorate = KClO3Decompostion: 2KClO3 ==> 3O2 + 2KCl7.5 moles KClO3 x 3 mole O2/2 moles KClO3 = 11.25 moles O2
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.
Reddish precipitate of Mercuric iodide and clear solution of Potassium chloride is produced
15 moles of O2 are produced by the decomposition if 10 moles potassium chlorate 2 KClO3 and 2 KCl plus 3O2.
potassium can make alot of compounds...i know a few... 1. potassium chloride (a healthier alternative to table salt/sodium chloride) 2. potassium nitrate 3. potassium hydroxide (produced when reacted with water)
The anwer is 0,061 moles of oxygen.
The reaction is:2 KClO3 = 2 KCl + 3 O238,4 moles of oxygen are produced.
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
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
The majority of the potassium chloride produced is used for making fertilizer, since the growth of many plants is limited by their potassium intake. As a chemical feedstock it is used for the manufacture of potassium hydroxide and potassium metal. It is also used in medicine, scientific applications, food processing, and as a sodium-free substitute for table salt (sodium chloride).
Hydrochloric acid evaporates off of potassium sulfate when it's produced. This results because potassium chloride is combined with sulfuric acid to create potassium sulfate.
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 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!!
The precipitate would be calcium carbonate, CaCO3.
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 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 reaction is:KOH + HCl = KCl + H2OThe answer is 5 moles KCl.
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
Balanced equation and potassium limits and drives the reaction.2K + Cl2 -> 2KCl6.75 grams K (1 mole K/39.10 grams)(2 mole KCl/2 mole K)(74.55 grams /1 mole KCl)= 12.9 grams potassium chloride produced==============================