using the equation calculate;what is the percent yield if you experimentally produce 3.65 grams of copper when 1.87 grams of aluminum reacts with 9.65 grams of copper(II) sulfate?plz help me with this questioni will be very thankful to you guysits about stoichiometry.
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Start with the balanced chemical equation: 2Al(s) + 3CuSO4(aq) - 3Cu(s) + Al2(SO4)3(aq) 1.87g .9.65g .3.65g Compute the mass of Cu from each of the masses of the reactants, assuming that each, in turn, is the limiting reactantThe smaller of the two masses of Cu will be the actual theoretical yieldDivide the 3.65, the actual yield, by the theoretical yield, x100, to get the percent yieldFor instance: If all of the aluminum reacted then you will get 6.60 g CuIf all of the copper (II) sulfate reacted then you will get 3.84 g CuThis tells us that copper (II) sulfate is the limiting reactant, and that we could no more than 3.84 grams of copper metalI will leave it to you to set up the conversion factors to get the masses of copper metalRemember, your teacher is expecting to see your workDivide to get the percent yield: 3.65 / 3.84 x 100 95.1%
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