The balanced chemical reaction would be as follows:
2H2O2 = 2H2O + O2
We are given the amount of the peroxide that decomposes. Using this as the starting point for the calculations, we can determine the amount of O2 produced. We do as follows:
14.3 mol H2O2 ( 1 mol O2 / 2 mol H2O2 ) = 7.15 mol O2 produced
The balanced equation is Fe₂O₃ + 3 CO = 2 Fe + 3 CO₂.
Next step is to convert everything to moles.
12.6g Fe₂O₃ x (1 mol Fe₂O₃ / 159.7g Fe₂O₃) = 0.07890 mol Fe₂O₃
9.65g CO x (1 mol CO / 28.01g CO) = 0.3445 mol CO
The third step is to determine the limiting and excess reactants.
0.07890 mol Fe₂O₃ x (3 mol CO/1 mol Fe₂O₃) = 0.2367 mol CO
Therefore Fe₂O₃ is the limiting reagent while CO is in excess.
0.07890 mol Fe x (2 mol Fe(s) / 1 mol Fe₂O₃) = 0.1578 mol Fe(s)
0.1578 mol Fe x (55.84g Fe / mole Fe) = 8.812g Fe is the theoretical yield
%yield = (7.23g / 8.812g) x 100% = 82.0% is the percent yield
The full question asks to decide whether the gas was a specific gas. That part is missing in your question. You need to decide whether the gas in the flask is pure helium.
To decide it you can find the molar mass of the gas in the flask, using the ideal gas equation pV = nRT, and then compare with the molar mass of the He.
From pV = nRT you can find n, after that using the mass of gass in the flask you use MM = mass/moles.
1) From pV = nRT, n = pV / RT
Data:
V = 118 ml = 0.118 liter
R = 0.082 atm*liter/mol*K
p = 768 torr * 1 atm / 760 torr = 1.0105 atm
T = 35 + 273.15 = 308.15 K
n = 1.015 atm * 0.118 liter / [ 0.082 atm*liter/K*mol * 308.15K] =0.00472 mol
mass of gas = mass of the fask with the gas - mass of the flasl evacuated = 97.171 g - 97.129 g = 0.042
=> MM = mass/n = 0.042 / 0.00472 = 8.90 g/mol
Now from a periodic table or a table you get that the molar mass of He is 4g/mol
So the numbers say that this gas is not pure helium , because its molar mass is more than double of the molar mass of helium gas.
