when 2.00g of copper(II) chloride dihydrate reacts with excess aluminum according to the following? 3CuCl2 ? 2H2O + Al ---> 3Cu + 2AlCl3 + 6H2Ohellppp
rephrase the question. I don't understand what you're asking for. Are you looking for the mass of something in particular? then tell me what it is you're looking for. First, convert 2g copper(II) chloride dihydrate into moles. That will give you the number of moles that are reacting, and you will therefore have how many moles of Cu(II), H2O, and Cl you will have (Cl and H2O will be multiplied by 2, and Cu will be multiplied by 3). from there, you should be able to get the number of moles of anything and therefore the mass of everything.
us the periodic table to find the molar mass of Cu and Cl. since Cu is 63.5, and Cl is 35.5, take the percent of Cu in the compound CuCl2 to find how many grams of Cu there are. 63.5 + 2(35.5) = 134.5 63.5 / 134.5 = 47.2 % this means, of the 2.00 grams, 47.2% of its mass is Cu, so:: .472 x 2.00 = .944 grams
The simple answer is that the moles of Cu produced will be the equal to the moles of dihydrate you start with. The balanced equation says that 3 moles of the hydrate reacts with 1 mole of Al to make 3 moles of Cu, 2 moles of AlCl3 and 6 moles of H2O. CuCl2*2H2O has a mass (rounded off) of 170.5 grams/mole. 2 g will be 2/170.5 = 0.01173 mole. So the Cu produced has to be 0.01173 mole = 0.01173 X 63.5 = 0.7448 grams
2g / GFM CuCl2 2g / ((35.453x2) + 63.546) Ans x (3Cu / 3CuCl2) Ans x 63.546 0.945259275 g of copper. (Ans means the answer of the previous expression, x means times)
3 moles copper(II) chloride dihydrate will produce 3 moles copper according to the balanced equation. 3 moles copper(II) chloride dihydrate = 3[ 64 + 2(35.5) + 2(18)] = 513g 3 moles copper = 3(64) = 192 grams 513 grams copper(II) chloride dihydrate will give 192 grams copper so 2 grams copper(II) chloride dihydrate = x cross multiplication; x = (2 x 192)/513 = 0.749 grams copper = 749 milligrams
