Theoretical yield

**Jonboy** · May 17th 2008, 06:00 PM

I think that the theoretical yield is the max amount of the products that can be formed. But I don't see how to find it.

The problem that is stumping me is this:

Given the equation $Cu_2 + O_2 \to 2Cu + SO_2$ , what is the theoretical yield of $SO_2$ if $8.20 g$ of $0_2$ are used ?

Will someone simply and clearly explain how to go about solving this?

**o_O** · May 17th 2008, 08:00 PM

You have 8.20 g of $O_{2}$ so we must "convert" it into terms of $SO_{2}$ . However, remember that a chemical reaction is representative of each species in moles. So that is where we will start off with:

$\text{8.20 g } O_{2} \times \frac{\text{1 mol } O_{2} }{\text{32.00 g }O_{2} } = \text{0.25625 mol } O_{2}$

Now, we know that for every mole of $SO_{2}$ produced, 1 mole of $O_{2}$ was used up. So we have a 1-1 ratio.

$\text{0.25625 mol } O_{2} \times \frac{\text{1 mol } SO_{2} }{\text{1 mol } O_{2} } = \text{0.25625 mol } SO_{2} \text{ produced}$

Note how I set up the ratio so that the units "mol $O_{2}$ " cancels out. That is pretty much the key in solving stoichiometry questions.

So you have the moles of $SO_{2}$ produced. Now all that is left is converting it into terms of grams and that should be your theoretical yield.

**Jonboy** · May 18th 2008, 06:03 AM

that makes complete sense! thank you.

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