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:
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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.
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Answer:
8.34
Explanation:
1) how much moles of NH₃ are in the reaction;
2) how much moles of H₂ are in the reaction;
3) the required mass of the H₂.
all the details are in the attachment; the answer is marked with red colour.
Note1: M(NH₃) - molar mass of the NH₃, constant; M(H₂) - the molar mass of the H₂, constant; ν(NH₃) - quantity of NH₃; ν(H₂) - quantity of H₂.
Note2: the suggested solution is not the shortest one.
