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

MnO₂ is limiting reactant

7.68g is theoretical yield

73.6% is percent yield

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 =

the law of conservation of mass states that in a chemical reaction, the total mass of reactants is equal to the total mass of products. because atoms are only rearranged in a chemical reaction, there must be the same number of sodium atoms and chlorine atoms in both the reactants and products

the potential energy of the reactants is 200j.

from the energy diagram, the energy of the product formed is 350j; this means that, this reaction is an endothermic reaction, because it absorbs energy from its environment.