What is the theoretical yield of MgO if 0.3645 grams of Mg react in excess O2 gas?

Question

What is the theoretical yield of MgO if 0.3645 grams of Mg react in excess O₂ gas?

Answer 1

Since the problem states that O2 is in excess, we know that Mg has to be the limiting reagent. Thus, we can use the amount of Mg given to determine the theoretical yield of MgO.

Balance the equation: 2 Mg + O₂ = 2 MgO : Mg -> MgO is 2:2 which is 1:1

According to the periodic table: 24.3g/mole is the atomic weight of Mg

Next, find the number of moles that 0.3645g Mg will produce:

0.3645g Mg = (1 mole Mg/ 24.3g Mg) = 0.0150 moles Mg

Since we know that Mg has a 1:1 mole ratio to MgO, then we know MgO will also produce 0.0150 moles.

Atomic weight of MgO = 24.3 MgO + 16 O₂ = 40.3g/mole MgO

0.0150 moles MgO x (atomic weight of MgO) =

0.0150 moles MgO x (40.3g/moles MgO) = 0.6045 grams of MgO