Answer to Question #167312 in Chemistry for Emily

Question

H2 can be generated from 15.38g of AI and 0.037 moles of NaOH when they react in excess water. What is the limiting reactant

Answer 1

Solution:

The balanced chemical equation:

2Al + 2NaOH + 2H₂O → 2NaAlO₂ + 3H₂

According to the equation above: n(Al) = n(NaOH)

Moles of NaOH = 0.037 mol

15.38 g Al × (1 mol Al / 26.981 g Al) = 0.570 mol Al

Choose one reactant and determine how many moles of the other reactant are necessary to completely react with it. Let's choose NaOH:

n(Al) = n(NaOH) = 0.037 mol

The calculation above means that we need 0.037 mol of Al to completely react with NaOH.

We have 0.570 mol of Al and therefore more than enough aluminium.

Thus aluminium (Al) is in excess and sodium hydroxide (NaOH) must be the limiting reagent.

Hence, the limiting reactant is sodium hydroxide (NaOH).

Answer: The limiting reactant is NaOH.