Question

please hurry! If 3.87g of powdered aluminum oxide is placed in a container containing 5.67g of water, what is the limiting reactant?

Answer 1

Answer:

Explanation:

Given parameters:

Mass of aluminium oxide = 3.87g

Mass of water = 5.67g

Unknown:

Limiting reactant = ?

Solution:

The limiting reactant is the reactant in short supply in a chemical reaction. We need to first write the chemical equation and convert the masses given to the number of moles.

Using the number of moles, we can ascertain the limiting reactants;

         Al₂O₃  + 3H₂O  →  2Al(OH)₃  

Number of moles;

            Number of moles = $\frac{mass}{molar mass}$

molar mass of Al₂O₃  = (2x27) + 3(16) = 102g/mole

    number of moles = $\frac{3.87}{102}$  = 0.04mole

   

molar mass of  H₂O = 2(1) + 16 = 18g/mole

    number of moles = $\frac{5.67}{18}$  = 0.32mole

From the reaction equation;

        1 mole of  Al₂O₃  reacted with 3 moles of H₂O

   0.04 mole of Al₂O₃ will react with 3 x 0.04 mole = 0.12 mole of H₂O

But we were given 0.32 mole of H₂O and this is in excess of amount required.

This shows that Al₂O₃ is the limiting reactant