Pure oxygen gas can be prepared in the laboratory by the decomposition of solid potassium chlorate to form solid potassium chloride and oxygen gas. 34.0 g
Nothing is produced, 500g potassium chlorate will be the same 500 g potassium chlorate after reaction. Actually there is no reaction at all.
5
(1.550M/liter of solution)(122.55 g/mole of KCLO3) = 190.0 g/liter(190.0 g/ 1000 mL) x 250.0 mL = 47.50 g/250.0 mL
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
224 g are in two moles of potassium dichromate.
Nothing is produced, 500g potassium chlorate will be the same 500 g potassium chlorate after reaction. Actually there is no reaction at all.
I Don't knows Sorry
Aproximately 4 grams of potassium chlorate will dissolve in 50 g of water at 20 degrees celsius.
5
(1.550M/liter of solution)(122.55 g/mole of KCLO3) = 190.0 g/liter(190.0 g/ 1000 mL) x 250.0 mL = 47.50 g/250.0 mL
224 grams of Oxygen will be in 2 moles of Potassium dichromate.
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
224 g are in two moles of potassium dichromate.
In one mole of potassium dichromate, there seven moles of oxygen. This means in two moles of K2Cr2O7, there are 14 moles of O, or 7 Moles of O2, which equals 224 grams.
The answer is 224,24 g oxygen.
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.
What is the mass, in grams, of 2 × 1012 atoms of potassium?