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 grams
50.0 grams KCl / (74.6 grams) = .670 moles KCl
To determine the number of moles of potassium iodide in 50 grams, you need to know the molar mass of potassium iodide, which is approximately 166 grams/mol. By dividing the given mass by the molar mass, you can calculate the number of moles. So, 50 grams of potassium iodide would be approximately 0.301 moles.
Since molecules of potassium contain only single potassium atoms, molecules of iodine contain two atoms, and moles of potassium iodide contain one atom of each element, 2.5 moles of iodine are needed to react completely with 5 moles of potassium.
To find the number of grams in 0.02 moles of beryllium iodide (BeI2), you would first calculate the molar mass of BeI2, which is 262.83 g/mol. Then, you would multiply the molar mass by the number of moles: 0.02 moles * 262.83 g/mol = 5.26 grams of beryllium iodide.
To find the number of moles of potassium iodide needed, multiply the volume of the solution (750 ml) by the molarity (1.8 moles/L). First, convert the volume to liters (750 ml = 0.75 L), then multiply 0.75 L by 1.8 moles/L to get 1.35 moles of potassium iodide.
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.
The molar mass of potassium carbonate (K2CO3) is approximately 138.21 g/mol. Therefore, 1.5 moles of potassium carbonate would be equal to 207.32 grams (1.5 moles * 138.21 g/mol).
530,3 g potassium iodide are needed.
Since molecules of potassium contain only single potassium atoms, molecules of iodine contain two atoms, and moles of potassium iodide contain one atom of each element, 2.5 moles of iodine are needed to react completely with 5 moles of potassium.
0.02 moles of beryllium diiodide = 5,256 grams
To find the number of grams in 0.02 moles of beryllium iodide (BeI2), you would first calculate the molar mass of BeI2, which is 262.83 g/mol. Then, you would multiply the molar mass by the number of moles: 0.02 moles * 262.83 g/mol = 5.26 grams of beryllium iodide.
34,7 moles of potassium 1 356,7 g.
To find the number of moles of potassium iodide needed, multiply the volume of the solution (750 ml) by the molarity (1.8 moles/L). First, convert the volume to liters (750 ml = 0.75 L), then multiply 0.75 L by 1.8 moles/L to get 1.35 moles of potassium iodide.
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
To calculate the grams of potassium permanganate in 2.20 moles, you would need to know the molar mass of potassium permanganate. The molar mass of potassium permanganate (KMnO4) is about 158.034 g/mol. So, 2.20 moles of KMnO4 is equal to 2.20 moles x 158.034 g/mol = 347.67 grams of potassium permanganate.
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.