The film obtained on the copper strip was a mixture of CuI and CuI2. What is the approx. mole ratio of CuI to CuI2 in the film, if a 4.000g strip of Cu reacted with the iodine vapor, given mass of 4.5000 g for the strip and film. After treatment with sodium thiosulfate, the dried strip weighed 3.8125g. PLEASE, HELP!!!
4.5000 - 4.000 = 0.5000 grams of iodide 0.5000 grams of iodide @ 126.90 g/mole = 3.940e-3 moles of Iodide 4.000 - 3.8125 = 0.1875 grams of copper 0.1875 grams of copper @ 63.546 g/mol = 2.951e-3 moles of copper each copper that reacted took at least one iodide... 3.940e-3 moles of iodide - 2.951e-3 moles of iodide for each copper = 0.989e-3 moles of Iodides left in excess for some coppers to have 2 iodides.... means that you made 9.89e-4 moles of CuI2 so how much, CuI was made: 2.951e-3 moles of copper - 9.89e-4 moles of CuI2 = 1.962e-3 CuI moles : 9.89e-4 moles of CuI2 1.962e-3 CuI mole ratio of CuI / CuI2 1.962e-3 CuI / 9.89e-4 moles of CuI2 = 1.98 the actual ratio is ....1.98 mole CuI to 1 mole CuI2 the approximate ratio os 2 mole CuI to 1 mole CuI2
