Question

For the following reaction, 6.02 grams of iodine are mixed with excess hydrogen gas . Assume that the percent yield of hydrogen iodide is 82.0 %.

hydrogen(g) + iodine(s) hydrogen iodide(g)

What is the theoretical yield of hydrogen iodide ?	grams
What is the actual yield of hydrogen iodide ?	grams

Answer 1

The chemical equation for the reaction between H219) and Izca) is as shown below, "tg) 1mde imde 2-T6) 2moles From the above balanced chemical equation, Imde of H₂ reacts with Imole of Iz to produce a moles of HI as product. * Molecular mass of Hq = 2xCH) = ax (Atomic mass of H] = &*(1.00784) = 2.01568 g/mole * Molecular mass of I, = 2x (Atomic mass do I) = 2x(1263.9.X = 253.808 g/mole * Molecular mass of HI = (Atomicinass 64-11)+(Atomic masiofi) = (1.00784) +(126.904) -127.911 g/mole H + 1200 " - 2 HI Imole imore =253.8089 = 2.01579 amoles = 2x (127-90) = 255.8229 From the above, 253,808 g of Iz produces 255.8229 with excess Hq gas.

Produces 253.8089.0f I, → 255.8229.0f HI 6.020.0 Iz produces a [(255-822) XC6.02) .. the amount of HI produced = (253.808) -6.06779 - " Theoretical yield of HI = 6.0679) * Percent yield = 1 / Actual yield ( Theoretical yield X100 Percent yield = 82.0% = (Actual yield), . 6.067) X100 :: Actual yields (82x6.0677 14.97559 100 Actual yield = 4.9755 g