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Stiochiometery

How many grams of magnesium(if any)remain when 5.00g of magnesium metal is burned in 2.50g of pure oxygen gas to give magnesium oxide?

Since this is clearly a homework problem, I'll guide you along. We're not here to just give you the answers to your homework.

First of all can you tell me the formula for Magnesium Oxide? How many magnesium atoms and how many oxygen atoms in each molecule of magnesium oxide? If you don't know, a 10 second Google search will tell you the answer.

Once you know how many of each atom there are in the molecule, you can find the ratio of the total weight of the two elements. For example, to find out how many AMUs (atomic mass units) of oxygen are in a magnesium oxide molecule, you'd multiply the number oxygen atoms per molecule times the atomic weight of oxygen.

The ratio of the weights of the two elements in one molecule will be the same as the ratio of the weights in any amount of the compound. Thus, you'll be able to write an equation setting the ratios equal to each other:

$\text{weight ratio in one molecule = weight ratio in any quantity}$

$\frac{\text{weight of magnesium in one molecule}}{\text{weight of oxygen in one molecule}} = \frac{\text{weight of magnesium in this problem}}{\text{weight of oxygen in this problem}}$

$\frac{\text{(No. of Mg atoms per molecule)} \; \times \; \text{(atomic weight of Mg)}}{\text{(No. of O atoms per molecule)} \; \times \; \text{(atomic weight of O)}} = \frac{x}{2.5\text g}$

Just solve for x, and that's how many grams of magnesium would be required to fully react with the 2.5 grams of oxygen. If the result is less than 5.00g, then clearly there's some left over. If it's greater than 5.00g, then all of the magnesium would get used up, and there wouldn't be any left over.

I form a equation for magnesium oxide
2Mg(s)+02(g)=2Mgo

stiochometric ratio
O2 Mgo
1 2
Mass: 2.5g ?
Atomic mass:32g/mol 40.3g/mol
Moles: 0.078125moles 0.078x2=0.15625moles

Mass of Mgo=0.15625x40.3=6.30g

Your math is okay, but you're taking the ratio of the wrong things.

It looks like you're finding the ratio of O2 to MgO. That's not helpful because MgO is the product of formation. You want the ratio of the two reactants, O to Mg (or Mg to O, either way) which form MgO.

As you can see just by looking at the chemical formula of magnesium oxide, MgO, you have 1 oxygen atom and 1 magnesium atom per molecule. Thus the weight ratio of Mg to O is

$\text{W.R. = }\frac{\text{weight Mg}}{\text{weight O}}=\frac{24.3}{16} = 1.52$

That's the same ratio of reactants that will be required when using 2.5g of oxygen. Thus,

$\frac{???\text{g Mg}}{\text{2.5g O}}=1.52$

Does that make sense?