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?

Answer 1

`"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"`.