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.
Calculate the grams of NaNO3 required to produce 5L of oxygen at STP if the percent yield of the reaction is 78.4%?
4 answers
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!
0.784 = 5 over x
then you divide 0.784 by 5?
I don't quite understand sorry!
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.
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.
Ah got it now thank you!