Question

A student reacts 1.109 g Al with 50 mL of 4 M NaOH and an excess of H2AO4 in solution to produce 16.58g of sodium alum.

-What is limiting reagent?

- What is theortical yield of sodium alum with a molar ration of 1:1?

Answer 1

The overall balanced reaction is:

Al + NaOH + H₂SO₄ -----------> NaAl(SO₄)₂·12H₂O

To know which is the limiting reagent, assuming a 1:1 relation as they stated in the problem, we need to get the moles of each reactant:

moles of Al = 1.109 g / 26.98 g/mol = 0.0411 moles

moles of NaOH = 4 mol/L * 0.050 L = 0.2 moles

1 mole of Al ------< 1 mole NaOH

0.0411 -----------> x

x = 0.0411 moles of NaOH, however we have 0.2 moles, so the limitant reactant is the Al

With this, the theorical yield would be:

m sodium alum = 0.0411 moles * (458.28 g/mol) = 18.84 g

Hope this helps.