# Chemistry

Calculate the grams of NaNO3 required to produce 5L of oxygen at STP if the percent yield of the reaction is 78.4%?

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1. 2NaNO3 ==> 2NaNO2 + O2
5L O2 at 78.4% yield would need to be 0.784 = 5/x and x = about 6.4
6.4L at STP = ? mols
mol = 6.4L/22.4L = approx 0.29
Convert mols O2 to mols NaNO3. That's 0.29mol O2 x 2 mol NaNO2/1 mol O2) = approx 0.57 mol NaNO3.
Then g NaNO3 = mols x molar mass
You should go through the problem and clean up the approximations.

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posted by DrBob222
2. Where does the 6.4 come from?
0.784 = 5 over x
then you divide 0.784 by 5?
I don't quite understand sorry!

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posted by Harry
3. You want 5L of O2. And if you could get 100% yield from the reaction you could calculate how much NaNO3 would be required to produce 5L O2. However, the problem says the reaction is only 78.4% yield so we must start with more NaNO2 so that at the end with just 78.4% yield we will have 5L O2. So I just divided 5L/0.784 = about 6.4L. In other words, if we want 5L O2 we must calculate how much NaNO3 to start with so it will produce 6.4L O2 so that 6.4 x 0.784 will give us the 5L we want.
yield = actual yield/theoretical yield
yield is 0.784. We want the actual yield to be 5L and I let x stand for theoretical yield. Then x = 6.4 and I didn't show that x is what we were solving for. To sum it up, we want 5L O2, the yield of the reaction is only 78.4% instead of 100%, so we must calculate g NaNO3 as if we were producing 6.4L O2.

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posted by DrBob222
4. Ah got it now thank you!

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posted by Harry

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