Additional information
Relative atomic mass(Ar) : A=7, G=16
The empirical formula : A₂G
<h3>Further explanation</h3>Given
3.5g of element A
4.0g of element G
Required
the empirical formula for this compound
Solution
The empirical formula is the smallest comparison of atoms of compound forming elements.
The empirical formula also shows the simplest mole ratio of the constituent elements of the compound
mol of element A :
mol of element G :
mol ratio A : G = 0.5 : 0.25 = 2 : 1
3940.2 is the volume of water vapour that would be produced from the combustion of 815.74 grams of propane () with 1,006.29 grams of oxygen gas, under a pressure of 1.05 atm and a temperature of 350. degrees C.
The ideal gas law (PV = nRT) relates the macroscopic properties of ideal gases. An ideal gas is a gas in which the particles (a) do not attract or repel one another and (b) take up no space (have no volume).
Stoichiometric calculations:
→
From the equation of the reaction, the mole ratio of propane to oxygen is 1:5.
Mole of 815.74 grams of propane =
Mole of 815.74 grams of propane = 18.49750567 moles
Mole of 1,006.29 grams of oxygen =
Mole of 1,006.29 grams of oxygen = 31.4465625 moles
Going by the mole ratio, it appears propane is limiting while oxygen is in excess.
From the equation, 1 mole of propane produces 4 moles of water vapour. Thus, the equivalent mole of water vapour will be:
18.49750567 moles x 4 = 73.99 moles.
Using the ideal gas equation:
PV = nRT
v = (73.99 x 0.08206 x 623) ÷ 0.96
v = 3940.2
Hence, 3940.2 is the volume of water vapour that would be produced from the combustion of 815.74 grams of propane () with 1,006.29 grams of oxygen gas, under a pressure of 1.05 atm and a temperature of 350. degrees C.
Learn more about the ideal gas here:
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