Now withidentified as the limiting reactan peohaced so the x(3morN) ng)-( (smo \

Question

Now withidentified as the limiting reactan peohaced so the x(3morN) ng)-( (smo ":) -3.33 mol NH3 pro duced OR 3 mol H2 2 mol NH3 5 mol Hr ) x-3.33 mol NH3 produced The answer from this calculation is called the theoretical yield, or the maximum amount of NH, that can be produced based on the starting materials and the reaction stoichiometry actually produced, is nearly always less than the theoretical yield for a number of possible reasons: the product is not totally recovered during the purification process; the reaction doesn't proceed completely due to reaction conditions; there might be evaporation of a volatile product such as a gas; or there may be a competing reaction (with its own chemical equation and stoichiometric relationships) where by-products are produced, thereby . The actual yield, the amount of product leaving less reactants available for the desired reaction. The percent yield (% yield) is a comparison of the actual to the theoretical yields and provides a measure of the efficiency of the reaction: actual yield theoretical yield % yield ×100 Equation 3 = THIS LAB In this lab, you will predict and observe a limiting reactant during the reaction between copper(II) chloride 3CuC12.2H,O(aq) + 2Al(s) 3Cu(s) + 2AIC,(aq) + 6H20 two metals as they change their oxidation states. CuCl, turns a light blue in aqueous solution due to the Cu2+(a) resulting reactions, you will be able to visualize limiting and excess reactants. You will also use stoichiometric dihydrate [CuCl, 2H,0) and metallic aluminum [AI(s)): This is known as an oxidation-reduction (or "redox") reaction, where electrons are exchanged between the ion, while aqueous AICl, is colorless. By varying the CuCl 2H,0 and Al(s) quantities and observing the 84 Laboratory 8 Limiting Reactant Lab 8-

Explanation / Answer

From the stoichiometry of the balanced chemical reaction, it can be observed that for every 3 moles of CuCl2 . 2H2O and 2 moles of Al(s), it will give 3 moles of Cu(s).

Molar mass of CuCl2 . 2H2O = 170.48 g/mol

Molar mass of 27Al = 27 g/mol

Analysis

(9) Beaker 1 calculations

mass of CuCl2 . 2H2O = 0.449 g

Moles of CuCl2 . 2H2O = 0.449 / 170.48

= 0.0026 mol

mass of Al = 0.244 g

moles of Al = 0.244 / 27

= 0.009 mol

Moles of Cu(s) produced (theoretical) = 0.0026 mol

mass of Cu(s) produced (theoretical) = 0.0026 * 63

= 0.1638 g

Mass of Cu(s) produced (actual) = 66.066 - 65.594

= 0.472 g

percenatge yield =

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