Question

2 Mg + 02(g) --> 2 MgOWhat is the excess reagent if 2.2 g of Mg is reacted with 4.5 L of oxygen at STP?

Answer 1

Answer: O2

Step-by-step explanation:

1) SOLVE FOR OPPOSITES

A) 2.2 g Mg x ( 1mole Mg / 24 g Mg) X 1 mole O2 / 2 mole Mg) X (22.4 L/1mole O2) =4.107 L O2

B) 4.5 x ( 1mole/ 22.4 L O2) X (2 mole Mg / 1 mole O2) X (24 g Mg/ 1 mole Mg) = 9.643 g Mg

2) now subtract answers from the number given ( negative number is limiting and positive number is excess)

2.2-9.643 = -7.443 + limiting reactant

4.5-4.107 = +0.393 L O2 and is the excess