This mass is 347,67 for KMnO4.
The compound potassium permanganate has chemical formula KMnO4 Molecular mass of KMnO4 = 39.1 + 54.9 + 4(16.0) = 158.0 Mass of KMnO4 = amount of KMnO4 x Molecular mass of KMnO4 = 2.55 x 158.0 = 403g
In 2 moles of potassium dichromate, there are 16 moles of oxygen atoms (from the two oxygen atoms in each formula unit). The molar mass of oxygen is 16 g/mol, so in 2 moles of potassium dichromate, there are 32 grams of oxygen.
You did not describe the amount of potassium bicarbonate amount in grams in your question. But if you are about 1 gram of potassium bicarbonate it will be 0.0099 moles in one gram of potassium bicarbonate. 0.0199 moles in 2 grams of potassium bicarbonate.
This is more of a math question that requires a bit of knowledge of chemistry. So it helps to know the steps of this answer mathematically. Additionally it's worth noting that there are a number of ways to answer this question. The method I provide may take an extra step, but it allows for a better understanding of the process.First we need to know some basic information about potassium permanganate, KMnO4. This basic information can be found on a periodic table, like the one in the link below. The first step is finding the weight of oxygen in one mole of potassium permanganate as a percent. For this you need to know the atomic weights of the elements involved.K: 39.1 gramsMn: 54.9 gramsO: 16.0 grams × 4 atoms = 64.0 gramsKMnO4: 39.1 + 54.9 + 64.0 = 158.0 grams/molSo now we know the weight of one mole of potassium permanganate (158.0 grams). Because we also know the weight of oxygen, we can find the percent of oxygen in the compound by mass.64.0 grams O ÷ 158.0 grams KMnO4 = 0.405 = 40.5%In one mole of potassium permanganate, 64.0 grams of it is oxygen, meaning 40.5% of it is oxygen. Because of the Law of Definite Proportions, we know that in any amount of potassium permanganate, 40.5% of it is oxygen.Then you can set up an equation.40.5% of (some amount of KMnO4) = 27.5 grams oxygenLet's set the amount of KMnO4 as the variable "x".0.405x = 27.5x = 67.9 grams KMnO4
The balanced equation for this reaction is: 2K3PO4 + 3Al(NO3)3 -> 6KNO3 + AlPO4. This indicates that 2 moles of potassium phosphate react with 2 moles of aluminum nitrate to produce 6 moles of potassium nitrate.
The compound potassium permanganate has chemical formula KMnO4 Molecular mass of KMnO4 = 39.1 + 54.9 + 4(16.0) = 158.0 Mass of KMnO4 = amount of KMnO4 x Molecular mass of KMnO4 = 2.55 x 158.0 = 403g
34,7 moles of potassium 1 356,7 g.
45/94.2 is 0.4777 moles
25,3 moles of potassium sulfate hva a mass of 4,4409 kg.
3.99 or 4
242.594 g
In 2 moles of potassium dichromate, there are 16 moles of oxygen atoms (from the two oxygen atoms in each formula unit). The molar mass of oxygen is 16 g/mol, so in 2 moles of potassium dichromate, there are 32 grams of oxygen.
Potassium permanganate = KMnO4 Molar mass of KMnO4 K = 1 * 39.10 g = 39.10 g Mn = 1 * 54.94 g = 54.94 g O = 4 * 16.00 g = 64.00 g Total = 158.04 g/mol 17.34 mol KMnO4 * (158.04 g/1 mol KMnO4) = 2740.41 g KMnO4 Convert the grams to kilograms. 1 kg = 1000 g 2740.41 g * (1 kg/1000 g) = 2.74041 kg Therefore, 17.34 moles of potassium permanganate is equal to about 2.74 kilograms.
For this you need the atomic (molecular) mass of KCl. Take the number of grams and divide it by the atomic mass. Multiply by one mole for units to cancel. KCl= 74.6 grams50.0 grams KCl / (74.6 grams) = .670 moles KCl
You did not describe the amount of potassium bicarbonate amount in grams in your question. But if you are about 1 gram of potassium bicarbonate it will be 0.0099 moles in one gram of potassium bicarbonate. 0.0199 moles in 2 grams of potassium bicarbonate.
Potassium has atomic number 39.1.Amount of K in 284g sample = 284/39.1 = 7.26molThere are 7.26 moles of potassium in a 284g sample.
To determine the grams of potassium chloride formed, you first need to calculate the moles of oxygen produced by the decomposition of potassium chlorate. Then, use the stoichiometry of the balanced chemical equation to convert moles of oxygen to moles of potassium chloride. Finally, from the molar mass of potassium chloride, you can calculate the grams formed.