Metallic aluminum reacts with mno2 at elevated temperatures to form manganese metal and aluminum oxide. a mixture of the two reactants is 67.2% mole percent al.

find the theoretical yield (in grams) of manganese from the reaction of 290g of this mixture.

Explanation:

To find the theoretical yield of the equation. First identify the limiting reactant in a chemical equation.

Step 1: write out the equation and balance it.

Al+ 3mno2=3mn+ 2Alo3.

The limiting reactant is mn02 because it is not found in excess.

Step 2: convert the % to gram . All contain 67.2% mole and mno2 will be 100-67.2= 32.8

All=67.2÷100×290(total gram of the reactants)=194.88g

Mno2=32.8÷100×290g=94.12g.

Step 3:calculate the molar mass of mno2 and that of mn. The atomic mass of mn is 54.9380 and that of oxygen is 16.

Mno2=54.938+ (16)2=86.98g/mol.

Mn=54.938.

Step 4:

From your balanced equation , calculate mn.

94.12g mno2× (1mol mno2÷86.98(molarmass) of mno2×3 mol of mn/4molAl×54.938g of mn÷1mol of mn.

94.12g×1÷86.98g×3÷4×54.938÷1

=44.58g

the answer is definitely c.