#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?

Question

1747 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-03-16T02:31:59

#"49.4 g of Mg"# reacted.

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"#.