Question

9.81g of manganese(iv) oxide react with 6.54 g of aluminum to make 5.65g of aluminum oxide and an unknown mass of manganese determine the limiting reagent theorectical yield and percent yielf

Answer 1

MnO₂ is limiting reactant

7.68g is theoretical yield

73.6% is percent yield

Step-by-step explanation:

The reaction of the question is:

3MnO₂ + 4Al → 2Al₂O₃ + 3Mn

To determine limiting reactant we need to convert each reactant to moles and, based on the chemical equation determine the moles you need of the other reactant for a complete reaction:

Moles MnO₂ -Molar mass: 86.937g/mol-:

9.81g * (1mol / 86.937g) = 0.113 moles

Moles Al -Molar mass: 26.98g/mol-:

6.54g * (1mol / 26.98g) = 0.242 moles

For a complete reaction of 0.242 moles of Aluminium you will require:

0.242mol Al * (3MnO₂ / 4Al) = 0.182 moles of MnO₂

As there are just 0.113moles of MnO₂, MnO₂ is limiting reactant

Now, with limiting reactant we need to determine theoretical moles and grams of Al₂O₃ produced:

Moles:

0.113 moles MnO₂ * (2Al₂O₃ / 3MnO₂) = 0.0753 moles Al₂O₃

Grams (Molar mass: 101.96g/mol):

0.0753 moles Al₂O₃ * (101.96g/mol) = 7.68g is theoretical yield

Percent yield is 100 times the ratio between actual yield (5.65g) and theoretical yield:

5.65g/7.68g * 100 =

73.6%