Calculate the grams of NaNO3 required to produce 5L of oxygen at STP if the percent yield of the reaction is 78.4%?
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.
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!
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.
Ah got it now thank you!
To calculate the grams of NaNO3 required to produce 5L of oxygen at STP, we need to use stoichiometry and consider the percent yield of the reaction.
1. Write the balanced chemical equation for the decomposition of NaNO3:
2NaNO3 → 2NaNO2 + O2
2. Determine the molar mass of NaNO3:
Na = 22.99 g/mol
N = 14.01 g/mol
O = 16.00 g/mol
Molar mass of NaNO3 = 22.99 + 14.01 + (3 * 16.00) = 85.00 g/mol
3. Determine the molar mass of O2:
O = 16.00 g/mol
Molar mass of O2 = 2 * 16.00 = 32.00 g/mol
4. Calculate the moles of O2 produced using the ideal gas law:
PV = nRT
At STP (standard temperature and pressure), the pressure (P) is 1 atm, the temperature (T) is 273.15 K, and the volume (V) is 5 L.
n = (PV) / (RT)
= (1 atm * 5 L) / (0.0821 L.atm/mol.K * 273.15 K)
≈ 0.2226 mol
5. Since the balanced equation shows that 2 moles of NaNO3 produce 1 mole of O2, we can calculate the moles of NaNO3 required:
moles of NaNO3 = (0.2226 mol O2) / 1
6. Consider the percent yield of the reaction (78.4%):
percent yield = (actual yield / theoretical yield) * 100
Rearranging the equation, we get:
actual yield = (percent yield / 100) * theoretical yield
In this case, the theoretical yield is the moles of NaNO3 calculated in step 5.
actual yield = (78.4 / 100) * (0.2226 mol NaNO3)
7. Finally, calculate the grams of NaNO3 by multiplying the actual yield in moles by the molar mass of NaNO3:
grams of NaNO3 = actual yield * molar mass of NaNO3
Just substitute the values from the previous calculations into this equation to find the answer.