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
50 g potassium iodide is equivalent to 0,3 moles.
0.388 moles
0.02 moles of beryllium iodide is equal to 5,256 g.
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.
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.
About 0.013
0.02 moles of beryllium iodide is equal to 5,256 g.
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
34,7 moles of potassium 1 356,7 g.
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.
224 g are in two moles of potassium dichromate.
25,3 moles of potassium sulfate hva a mass of 4,4409 kg.
45/94.2 is 0.4777 moles
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.
3.99 or 4
About 0.013