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!!
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
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.
2 to 3, because of the balanced equation:2 KClO3 --> 2 KCl + 3 O2
there are two moles produced in potassium nitrate.
Balanced equation. 4Na + O2 -> 2Na2O 10 moles Na (2 moles Na2O/4 moles Na) = 5.0 moles Na2O produced
12 moles KClO3 (3 moles O/1 mole KClO3) = 36 moles of oxygen.
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.
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
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.
Mole percent, or molar percent of a substance is the ratio of the moles of a substance in a mixture to the moles of the mixture. It represents the number of moles of a substance in a mixture as a percentage of the the total number of moles in the mixture. Mole % = (mol substance in a mixture) / (mol mixture) * 100
2 KClO3 ------ 2KCl + 3O2 so 2 moles of KClO3 produces two mole of KCl. Therefore 0.440 moles of potassium chlorate will produce 0.44 moles of KCl - potassium chloride.
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
2KClO3==>2KCl+3O2 is the equation. so you need 4 moles of KClO3.
2 grams of Oxygen can be obtained from 5 grams of KClO3 (only if the "CL" means "Cl", which is Chlorine! Remember that only the first letter of the atomic symbol is capitalized.)
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
2 to 3, because of the balanced equation:2 KClO3 --> 2 KCl + 3 O2
3 moles of oxygen are obtained from 2 moles of potassium chlorate.M of KClO3 is 122,55 g, M of O2 is 32, density of oxygen is 1,429 g/L.