#Mg# and #O_2# react in a 2.1 molar ratio. 2 moles #M_g# = 1 mole #O_2#. If a reaction used 32.5 g of #O_2#, how many g of #Mg# reacted?

1 Answer
Mar 16, 2016

Answer:

#"49.4 g of Mg"# reacted.

Explanation:

Step 1. Calculate the moles of #"O"_2#.

#32.5 cancel("g O"_2) × ("1 mol O"2)/(32.00 cancel("g O"2)) = "1.016 mol O"_2#

Step 2. Use the molar ratio to calculate the moles of #"Mg"#.

#1.016 cancel("mol O"_2) × "2 mol Mg"/(1 cancel("mol O"_2)) = "2.031 mol Mg"#

Step 3. Calculate the mass of #"Mg"#.

#2.031 cancel("mol Mg") × ("24.30 g Mg")/(1 cancel("mol Mg")) = "49.4 g Mg"#

The reaction required #"49.4 g Mg"#.