Question

LT.3b - STOICHIOMETRY - QUESTION 3Magnesium burns in air according to the equation below. Howmany grams of oxygen gas are needed to completely burn 4.51moles of magnesium?Mg + O2->MgOA. 72.2 gB. 144 gC. 36.1gD. 289 g

Answer 1

A. 72.2 g

Step-by-step explanation

Given:

Moles of Mg = 4.51 mol

Reaction: Mg + O₂ ----> MgO

What to find:

The grams of oxygen gas needed to completely burn 4.51 moles of Mg.

Step-by-step solution:

The first step is to balance the given chemical reaction.

Mg + 1/2O₂ ----> MgO

The next step is to calculate the number of moles of oxygen gas needed.

From the balanced reaction; 1 mol Mg reacts with 1/2 mol O₂

So, 4.51 mol Mg will react with:

`\frac{4.51\text{ }mol\text{ }Mg*(1)/(2)\text{ }mol\text{ }O_2}{1\text{ }mol\text{ }Mg}=2.255\text{ }mol\text{ }O_2`

The final step is to calculate grams of oxygen in 4.5 moles of oxygen that was needed.

1 mole of O₂ = 32 grams

Therefore, 2.255 moles of O₂ = (2.255 x 32 grams) = 72.16 g = 72.2 g

The correct answer is:

A. 72.2 g