Answer to Question #168374 in Chemistry for Emily

Question

In the following reaction 12.5g of Al and 7.6 g of NaOH react in excess water. Calculate the number of moles Al and NaOH before the reaction occurs, determine the limiting and the excess reactant. How many moles would be left over of the excess reactant

Answer 1

2Al + 6NaOH = 2Na₃AlO₃ + 3H₂

n = m / M

M (NaOH) = 40 g/mol

M (Al) = 27 g/mol

n (Al) = 12.5 / 27 = 0.5 mol

n (NaOH) = 7.6 / 27 = 0.3 mol

According to the reaction, 2 moles of Al requires 6 moles of NaOH (the ratio is 2:6 or 1:3).

Therefore, the available 0.5 moles of Al require 0.5x3=1.5 moles of NaOH.

So that NaOH is the limiting reactant in this case. It is used fully. The remaining amount of Al is:

0.5 - 0.3/3 = 0.4 moles