limiting reagent problem

Question

Pure magnesium metal is often found as ribbons and can easily burn in the presence of oxygen. When 4.69 g of magnesium ribbon burns with 8.72 g of oxygen, a bright, white light and a white, powdery product are formed. Enter the balanced chemical equation for this reaction. Be sure to include all physical states.

The percent yield for the reaction is 80.7%, how many grams of product were recovered?

How many grams of the excess reactant remain?

Answer 1

Ok, so lets think about this question, and everything we need to figure out to answer the question.

First, we need to figure out what reaction is happening, and we need to make sure we balance that equation.

Then we need to determine the limiting reagent.

Finally we need to determine how much product we could theoretically make, as 80.7% of that amount represents out actual yield.

So starting from the beginning, Mg likes to take on a +2 charge because it is in the second column of the periodic table. Oxygen likes to take on a -2 charge. So what salt are we making in the end? Well, lets assume the Mg and O both take on the charges they are happy with. So Mg²⁺ and O^2-, combine and become a compound with balanced charge. since their charges are the same magnitude but in opposite directions it would just be a 1:1 compound (MgO is the product)

So we know:

Mg + O₂ --> MgO

but this is not yet balanced, we only have 1 oxygen in the products and 2 in the starting materials.

So we just need to consume the remaining oxygen by using another Mg:

2Mg + O₂ --> 2MgO

Now, the equation is balanced.

So now we figure out the limiting reagent,

The easiest way to do limiting reagent problems is just to figure out how much product each reactant allows you to make. Whichever reactant lets you make the smallest amount of product is the limiting reagent.

Now we have 8.72g O_2, and O₂ is 32 g/mol, so we have 0.2725mol of O₂ (8.72g / 32g/mol = .2725mol). For every mole of O₂ we can make 2 moles of MgO (see balanced equation above), meanong our O₂ would allow us to make 0.545moles of MgO.

We also have 4.69g Mg. Mg is 24.3 g.mol, so we have 4.69g / 24.3g/mol = .193moles of Mg. For every mole of Mg, we can make 1 mole of MgO, so based on Mg we can make 0.193 moles of MgO.

0.193 < .545, so Mg is the limiting reagent. This also means the greatest amount of product we could theoretically make is .193 moles. However, our actual yield is 80.7% of this. so .193moles *0.807 = .156 moles. We made 0.156 moles of product. All we need now is to figure out the mass of MgO represented by 0.156 moles. So 0.156* the molar mass of MgO (40.3g.mol) = the mass of product recovered (6.27g MgO)

So all thats left is how much excess O₂ remains. O₂ was the reagent in excess, so the second part of the question refers to the remaining O₂. We know we ended up making 0.156 moles of MgO, meaning we must have consumed .078 moles of O2 (because 1 mole of O₂ makes 2 moles of MgO). We started with 0.2725 moles of O₂, so we're left with 0.2725 - .0.078 = 0.1945moles of O₂. The molar mass of O₂ is 32, so 0.1945moles * 32g/mol = 6.224g O₂ remaining after the reaction.

If you would like to go through some practice problems, or review any other concepts please feel free to contact me.

Answer 2

Mg(s) +O₂(g) ==> MgO(s)

moles Mg present = 4.69 g x 1 mole/24.3 g = 0.193 mol Mg

moles O₂ present = 8.72 g x1 mol/32 g = 0.273 mol O₂

Limiting reactant is Mg

Theoretical yield of MgO = 0.193 mol Mg x 1 mol MgO/mol Mg x 40.3 g/mol MgO = 7.78 g

Mass MgO recovered = 0.807 x 7.78 g = 6.28 g

For O₂, 0.273 mol - 0.193 mol = 0.08 mol remaining

Mass O₂ remaining = 0.08 moles x 32 g/mole = 2.56 g