For the following reaction, 3.00 grams of chlorine gas are mixed with excess sod

Question

For the following reaction, 3.00 grams of chlorine gas are mixed with excess sodium iodide. The reaction yields 4.38 grams of sodium chloride.


chlorine (g) + sodium iodide (s) sodium chloride (s) + iodine (s)
% yield=
theoretical=

Explanation / Answer

chlorine (g) + sodium iodide (s) sodium chloride (s) + iodine (s) % yield= theoretical= Cl2(g) + 2 NaI (s) =======> 2 NaCl + I2 Moles = mass/molar mass Moles of Cl2 = 3.00g / 70.91 g/mol =0.0423 mol Cl2 Now the molar ratio is 1 Cl2: 2 NaCl 0.0423 mol Cl2 * ( 2 mol NaCl / 1mol Cl2)*(58.44 g NaCl/ 1 mol NaCl) 4.94 g NaCl===> Theoretical yield % yield = (actual yield/theoretical yield)*100 = (4.38/4.94)*100 = 88.86%

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