The balanced reaction: C2H6 + 7/2 O2 -> 2 CO2 + 3 H2O
We first convert volume of C2H6 to no. of moles. We use the conditions at STP where 1 mol = 22.4 L thus,
Moles C2H6 = 16.4 L/ 22.4 L =0.7321 mol
In order to determine the limiting reagent, we look at the given amounts of the reactants.
0.7321 mol C2H6 (7/2 mol O2 / 1 mol C2H6) = 2.562 mol O2
From the given amounts of the reactants, we can say that O2 is the limiting reactant since we need 2.562 mol O2 to completely react the given amount of C2H6. The excess reagent is C2H6
To calculate for the amount of products and excess reactants:
0.980 mol O2 (2 mol CO2 / (7/2 mol O2)) = 0.56 mol CO2 (22.4 L / 1 mol ) =12.544 L CO2
<span>0.980 mol O2 (1 mol C2H6 / (7/2 mol O2)) = 0.28 mol C2H6
Excess C2H6 = 0.7321 mol - 0.28 mol C2H6 = 0.4521 mol C2H6
We then use the molecular weight of C2H6 to convert the excess amount to grams.
0.4521 mol C2H6 (30.08 g C2H6 / 1 mol C2H6) = <span>13.60 g C2H6
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<span>Since the limiting reagent is O2 there will be no oxygen atoms that will be left after the reaction.</span>
The volume of 0.125 mole of the gas at the same temperature and pressure 0.30 L
<h3>Data obtained from the question </h3>
- Initial mole (n₁) = 0.82 mole
- Initial volume (V₁) = 2 L
- New mole (n₂) = 0.125 mole
- Pressure = Constant
- Temperature = Constant
- New Volume (V₂) =?
<h3>How to determine the new volume</h3>
The new volume can be obtained as illustrated below
PV = nRT
Divide both side by P
V = nRT / P
Divide both side by n
V / n = RT / P = Constant
Thus,
V₁ / n₁ = V₂ / n₂
2 / 0.82 = V₂ / 0.125
Cross multiply
0.82 × V₂ = 2 × 0.125
Divide both side by 0.82
V₂ = (2 × 0.125) / 0.82
V₂ = 0.30 L
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There is 55 neutrons in Cesium-130.
Answer:
The Bohr Model is very limited in terms of size. Poor spectral predictions are obtained when larger atoms are in question. It cannot predict the relative intensities of spectral lines. It does not explain the Zeeman Effect, when the spectral line is split into several components in the presence of a magnetic field.
Explanation:
