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.

<h3>What is an ideal gas equation?</h3>

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:

brainly.com/question/27691721

#SPJ1

7 0

2 years ago

How many grams of water will be produced from 50 g hydrogen reacting with 50 g oxygen?

DanielleElmas [232]

We need to first come up with a balanced equation:

$4H+O_{2}$ → $2H_{2}O$

We know that the molar ratio of hydrogen to oxygen to water now is 4:1:2.

Converting the amount of grams given to moles is as follows:

Hydrogen: $\frac{1mole}{1.008g}*50g=49.6mol$

Oxygen: $\frac{1mole}{15.999g}*50g=3.125mol$

We know now that the limiting reactant is oxygen. We can then know that the number of moles of water are produced are double the number of moles of oxygen used due to the ratio that we established at the beginning - 4:1:2.

So we now can use 6.25 moles of water as the amount produced.

Then we convert moles of water to grams:

$\frac{18.015g}{1mole} *6.25mol=112.59g$