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.)

Answer 1

Molar mass of Al = 27 g/mol
mass of Al = 0.143 g
Number of moles of Al = mass/ molar mass
                                        = 0.143 /27
                                        = 0.0053 mol

number of moles of HNO3 = M*V
                              = 0.5 M* 0.2L
                              =0.01 mol

From reaction, 2 mol of Al reacts with 6 mol of HNO3.
0.0053 mol of Al will react with = 6*0.0053/2 =0.0159 mol of HNO3
But we have only 0.01 mol of HNO3.
So HNO3 is limiting reagent.
We will use HNO3 in our further calculations.
6 mol of HNO3 gives 3 mol of H2

So,
Theoretically, number of moles of H2 produced = 3*0.01/6 =0.005
Molar mass of H2 = 2g/mol
Mass of H2 = number of moles * molar mass
                    =0.005*2
                    = 0.01 g   (<-----This is theoretical yield)

Actual produced = 0.00842 g
Percent yield = ACtual yield * 100 / theoretical yield
                         =0.00842*100/0.01
                         =84.2 %