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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
A How many moles of O2 were produced b What percent of the mixture was KClO3 and KCl?
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!!
When 40 grams of ammonium nitrate explode 14 grams of nitrogen and 8 grams of oxygen form how many grams of water form?
the same amount would have to stay in grams, so if 14 grams of nitrogen is formed, then 8 grams of oxygen, add those two together and you get 22. and that's 22 of the 40 grams used, so 40 subtracted by 22 is 18. 18 grams of water would be formed.
Calculate the number of moles of oxygen atom in 18.1 mole of CH2O?
Since there is only 1 oxygen atom in CH2O, there is the same amount of oxygen atoms as there are molecules of CH2O. So the answer is 18.1 mole. But if you burn it, you will form oxygen in its natural form, which is O2. So there will only be the half the amount of the oxygen. Then the answer would be 18.1 / 2 = 9.05 mole.
How many grams of oxygen are in 5.75 moles of aluminum oxide?
Well to find how many grams are in moles you would eventually multiply the mole by the molar mass. The molar mass of aluminum oxide would be 101.96 ( you would find that by multiplying the atomic mass of al by 2 and o by 3 and adding them together). But the molar mass of Oxygen is just about 48 (rounded to 16 instead of 15.9994)5.75 moles of Al2O3 X 48 g oxygen/1 mole of Al2O3=276 g oxygen in 5.75 mole Al2O3
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 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 KClO3 on decomposition give 1 mole of oxygen?
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.)
A How many moles of O2 were produced b What percent of the mixture was KClO3 and KCl?
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!!
2 H2 plus O2 2 H2O - How many moles of water will be produced from a fuel cell that consumes 0.633 moles of oxygen?
For every mole of oxygen consumed in the reaction 2H2 + O2 -> 2H2O, two moles of water are produced. Therefore, if 0.633 moles of oxygen are consumed, the number of moles of water produced would be 2 x 0.633 = 1.266 moles.
How many moles of oxygen will be produced from the complete combustion of 4 moles of propane?
The balanced chemical equation for the combustion of propane is: C3H8 + 5 O2 -> 3 CO2 + 4 H2O. This means that 5 moles of oxygen are required to completely combust 1 mole of propane. Therefore, 20 moles of oxygen would be produced from the complete combustion of 4 moles of propane.
What is the grams of substance in 2 moles of KClO3?
Alls you do to find a molar mass is add up all of the atomic masses. Potassium=39.09 Chlorine= 35.453 Oxygen(3)=15.999. So KClO3 would equal 271.65g to a mol? Then I think to find the number of atoms of each you would take the atomic mass * avacodo's number so it would be for example chlorine. 1gCl(35.453/1)(6.022*10^23/1) However I might be wrong on that part.
If 4.8g of oxygen gas is completely converted to ozone how many molecules of ozone would be produced?
4.8/16 moles of oxygen atoms converts to 1.6/16 moles of ozone molecules.
In the reaction 2CO plus O2 equals 2CO2 what is the ratio of moles of oxygen used to moles of CO2 produced?
If you mean atoms then two, if molecules one.
When 36.3 g of potassium chlorate is heated 22.1 g of potassium chloride is produced What mass of oxygen is produced?
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.
How many moles of mercury are produced if 125 g of oxygen is also produced?
To determine how many moles of mercury are produced when 125 g of oxygen is generated, we first need to know the balanced chemical equation for the reaction involving mercury and oxygen. Assuming the reaction is the formation of mercury(II) oxide (HgO) from mercury (Hg) and oxygen (O₂), the equation is: 2 Hg + O₂ → 2 HgO. Given that the molar mass of oxygen (O₂) is approximately 32 g/mol, 125 g of oxygen corresponds to about 3.91 moles of O₂. According to the stoichiometry of the balanced equation, 2 moles of Hg are produced for every 1 mole of O₂. Therefore, 3.91 moles of O₂ would produce approximately 7.82 moles of Hg.
How many moles of SO3 would be produced when 1.2 moles of oxygen is consumed in the equation S plus O2 equals SO3?
First you need to find the balanced reaction:2S + 3O2 --> 2SO3So using the balanced reaction we see that for every 3 moles of oxygen consumed, 2 moles of sulfur trioxide are produced:1.2 moles O2 consumed * (2 moles SO3/3 moles O2) = 0.8 mole of SO3 produced
