Question

What volume of water vapor would be produced from the combustion of 815.74 grams of propane (C3H8) with 1,006.29 grams of oxygen gas, under a pressure of 1.05 atm and a temperature of 350. degrees C? Given:
C3H8(g) + 5 O2(g) ---> 3 CO2(g) + 4 H2O(g)

(OR C3 H8 ("g") + 5 O2 ("g") right arrow 3 C O2 ("g") + 4 H2O ("g")




Do not type units with your answer

Answer 1

3940.2 is the volume of water vapour that would be produced from the combustion of 815.74 grams of propane ($C_3H_8$) with 1,006.29 grams of oxygen gas, under a pressure of 1.05 atm and a temperature of 350. degrees C.

What is an ideal gas equation?

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:

$C_3H_8(g) + 5 O_2(g)$→ $3 CO_2(g) + 4 H_2O(g)$

From the equation of the reaction, the mole ratio of propane to oxygen is 1:5.

Mole of 815.74 grams of propane = $\frac{ 815.74}{44.1 }$

Mole of 815.74 grams of propane = 18.49750567 moles

Mole of  1,006.29 grams of oxygen =$\frac{ 1,006.29}{32 }$

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 ($C_3H_8$) 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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