Question

Given the reaction: 2 Al (s) + 6 HNO₃ (aq) ® 2 Al(NO₃)₃ (aq) + 3 H₂ (g) .. When 0.143 g of Al were added to 200 mL of 0.500 M HNO₃, 0.00842 g of H2 were produced. Find the percent yield of H₂.

(Please don’t forget to calculate the theoretical yield of hydrogen from both the aluminum and the nitric acid.)

the answer should be = 53.0%

Answer 1

moles of Al = 0.143 / 27 = 0.0053

moles of HNO3 = 200 x 0.5 /1000 = 0.1

moles of H2 = 0.00842 / 2 = 0.00421

2 Al (s) + 6 HNO₃ (aq) ------------------> 2 Al(NO₃)₃ (aq) + 3 H₂ (g)

2 mol          6 mol

0.0053        0.1

here limiting reagent is Al. so the product is based on Al

2 mol of Al produce -------------- 3 mol of H2

0.0053 mol Al           ---------------   x mol H2

moles of H2 = 0.0053 x 3 / 2

                    = 0.00795

mass of H2 = moles x molar mass

                    = 0.00795 x 2

                     = 0.0159 g

theoretical yield of hydrogen = 0.0159 g

actual yield = 0.00842 g

% yield = (actual / theoretical) x 100

             = (0.00842/0.0159) x 100

             = 53 .0 %