#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

#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?

Naomi Aronson · Accepted Answer

#"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"#.

Xavier Avery · Answer

undefined First, determine the number of moles of O2 used in the reaction using its molar mass.

Molar mass of O2 = 16.00 g/mol (for each oxygen atom)
So, the molar mass of O2 = 16.00 g/mol * 2 = 32.00 g/mol

Given: Mass of O2 = 32.5 g

Now, calculate the number of moles of O2 used:
Number of moles of O2 = Mass of O2 / Molar mass of O2

Then, apply the molar ratio between Mg and O2 to find the number of moles of Mg reacted.
Given: Molar ratio of Mg to O2 is 2:1

So, if 1 mole of O2 reacts, 2 moles of Mg react.

Finally, use the calculated moles of O2 and the molar ratio to find the moles of Mg reacted. Then, convert the moles of Mg to grams using its molar mass.

This process will give you the mass of Mg that reacted.