Here is my question:How many grams of copper should be produced when 2.00g of copper (II) chloride dihydrate reacts with excess aluminum according the following equation?3CuCl2 o 2H2O + 2Al --> 3Cu +2AlCl3 +6H2OMy answer was 0.878g. Is this correct?
First you need to find the percentage of the mass of copper (ii) chloride dihydrate that is only copper. This can be found with the molecular/atomic weights. Cu - 63.54 Cl - 35.453 * 2 = 70.906 H2O - 18 * 2 = 36 Total for the mixture = 170.446 % Cu = 37.28% Mass Cu = 0.7456 g Cu cannot be created nor destroyed and Cu does not appear anywhere in the products except as elemental Cu. The answer should be 0.7456 g.
2.00 g * mol / 170.48 g * 3 mol / 3 mol * 63.55 g / mol = dyhydrus CuCl2 has atomic mass of 170.48 copper hass atomic mass of 63.55 2 * 63.55 / 170.48 = .746 g Cu made
