Answer to Question #169355 in Chemistry for Emily

Question

How many grams of H2 can be generated from 17.25g of Al and 0.947 moles of NaOH react in excess water? Determine the limiting and the excess reactants? How many grams of H2 are formed? In the lab 1.72 g of H2 was formed, what is the percent yield

Answer 1

In water, the reaction is as follows:

2Al+2NaOH+6Н₂О = 2Na₃[Al(OH)₆] + 3H₂

The number of moles of Al in the reaction:

n(Al) = m(Al) / [2 × Mr(Al)] = 17.25 g / (2 × 26.98 g/mol) = 0.32 mol

Since n(Al) < n(NaOH), Al is a limiting reactant. From here:

m(H₂) = n(Al) × 3 × Mr(H₂) = 0.32 mol × 3 × 2 g/mol = 1.92 g

The percent yield equals:

%w = (actual yield / theoretical yield) × 100% = (1.72 g / 1.92 g) × 100% = 89.6 %