A 0.532 mol sample of SO2 gas requires 52.3 s to effuse through a tiny hole. Under the same conditions, how long will it take 0.
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Answer:
The molecular formula =
Explanation:
% of C = 11.79
Molar mass of C = 12.0107 g/mol
<u>% moles of C = 11.79 / 12.0107 = 0.9816</u>
% of Cl = 69.57
Molar mass of Cl = 35.453 g/mol
<u>% moles of Cl = 69.57 / 35.453 = 1.9623</u>
Given that the gaseous chlorofluorocarbon only contains chlorine, flourine and carbon. So,
% of F = 100% - % of C - % of C = 100 - 11.79 - 69.57 = 18.64
Molar mass of F = 18.998 g/mol
<u>% moles of F = 18.64 / 18.998 = 0.9812</u>
Taking the simplest ratio for C, Cl and F as:
0.9816 : 1.9623 : 0.9812
= 1 : 2 : 1
The empirical formula is =
Also, Given that:
Pressure = 21.3 mm Hg
Also, P (mm Hg) = P (atm) / 760
Pressure = 21.3 / 760 = 0.02803 atm
Temperature = 25 °C
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
So,
T = (25 + 273.15) K = 298.15 K
Volume = 458 mL = 0.458 L (1 mL = 0.001 L)
Using ideal gas equation as:
PV=nRT
where,
P is the pressure
V is the volume
n is the number of moles
T is the temperature
R is Gas constant having value = 0.0821 L.atm/K.mol
Applying the equation as:
0.02803 atm × 0.458 L = n × 0.0821 L.atm/K.mol × 298.15 K
⇒n = 0.00052445 moles
Given that :
Amount = 0.107 g
Molar mass = ?
The formula for the calculation of moles is shown below:
Thus,
Molecular formulas is the actual number of atoms of each element in the compound while empirical formulas is the simplest or reduced ratio of the elements in the compound.
Thus,
Molecular mass = n × Empirical mass
Where, n is any positive number from 1, 2, 3...
Mass from the Empirical formula = 1×12.0107 + 2×35.453 + 1×18.998 = 101.9147 g/mol
Molar mass = 204.0233 g/mol
So,
Molecular mass = n × Empirical mass
204.0233 = n × 101.9147
⇒ n = 2
<u>The molecular formula = </u>
Answer : The theoretical yield of HCl is, 56.1735 grams
Explanation : Given,
Mass of = 60 g
Mass of = 37.5 g
Molar mass of = 117 g/mole
Molar mass of = 18 g/mole
Molar mass of = 36.5 g/mole
First we have to calculate the moles of and
.
Now we have to calculate the limiting and excess reagent.
The balanced chemical reaction is,
From the balanced reaction we conclude that
As, 1 mole of react with 3 mole of
So, 0.513 moles of react with
moles of
From this we conclude that, is an excess reagent because the given moles are greater than the required moles and
is a limiting reagent and it limits the formation of product.
Now we have to calculate the moles of .
As, 1 mole of react to give 3 moles of
So, 0.513 moles of react to give
moles of
Now we have to calculate the mass of .
Therefore, the theoretical yield of HCl is, 56.1735 grams