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Sep 21 2015 04:06pm
Consider the reaction of sulfur (S) and oxygen gas (O2) to give sulfur dioxide.

Double the mass of oxygen as compared to sulfur was used during a reaction, which produced the same mass of sulfur dioxide as the total mass of oxygen.

What is the percent yield of the reaction?
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Sep 21 2015 05:12pm
This is basic stoich chemistry. The first thing you should usually do is write out a balanced equation. This will tell you how many moles of each reactant (sulfur and oxygen) it takes to produce however many moles of the product (sulfur dioxide)

After that, you'll need to consider the molar masses of all of them. Say you had x grams of sulfur, and 2x of both oxygen and sulfur dioxide. Use that to figure out how many moles of each you had (in terms of x). Looking at the balanced reaction, decide which of the two will run out first (limiting reactant). Let's say the reaction consumes 2 moles of sulfur and 1 mole of oxygen (it doesn't really), and you had 3 moles of sulfur and 2 moles of oxygen. You will run out of sulfur first, because consuming 3 moles of sulfur will only consume 1.5 moles of oxygen.

Once you know the limiting reactant, calculate how many moles of the product you would produce with it.

You already calculated how many you actually produced earlier (using the fact that you had 2x mass)

Finally, divide the amount you actually got by the amount you should have gotten, and you'll get your percent yield (the x variable will cancel out).

This post was edited by russian on Sep 21 2015 05:17pm
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