Question

If 84.1 g of NaOH and 51.0 g of Al and 25.0 g H20 react which chemical is the limiting reactant?

Answer 1

Answer:

• H₂O

Explanation:

1. Chemical quation

The reaction of aluminium, sodium hydroxide and water is represented by the balanced chemical equation:

• 2Al(s) + 2NaOH(s) + 6H₂O(l) → 2Na[Al(OH)₄] (aq) + 3H₂(g) ↑

The coefficients of each reactant and product give the theoretical mole ratios.

To find the limiting reactant you compare the theoretical ratios with the ratio of the available substaces.

2. Theoretical mole ratio:

• 2 mol Al : 2 mol NaOH : 6 mol H₂O

Equivalent to

• 1 mol Al : 1 mol NaOH : 3 mol H₂O

3. Actual ratio

a) Convert each mass to number of moles

Formula:

• number of moles = mass in grams / molar mass

Al:

• molar mass = atomic mass = 26.982g/mol

• number of moles = 51.0g / 26.982g/mol = 1.89 mol

NaOH:

• molar mass = 39.997g/mol

• number of moles = 84.1g / 39.997g/mol = 2.10 mol

H₂O:

• molar mass = 18.015g/mol

• number of moles = 25.0g / 18.015g/mol = 1.39 mol

Divide all the mole amounts by the least number:

• Al: 1.89/1.39 = 1.36
• NaOH: 2.10 = 1.52
• H₂O: 1.39 = 1.00

• 1.36 mol Al : 1.52 mol NaOH : 1.00 mol H₂O

4. Comparison

Theoretical ratio:

• 1 mol Al : 1 mol NaOH : 3 mol H₂O

Actual ratio:

• 1.36 mol Al : 1.52 mol NaOH : 1.00 mol H₂O

     Multiply by 3:

• 4.08 mol Al : 4.56 mol NaOH : 3.00 mol H₂O

Now, yo can see that the first two are in excess with respect the third one, making that the water consumes first, before any of the other two consumes. Therefore, the limiting reactant is the water.