Question

In a gas grill, 29 lbs propane C3H8 are

burned with just enough air for complete combustion at a party. How
many lbs of combustion products are formed? Round your answer to
the nearest whole number.

Answer 1

Answer : The mass of combustion products formed are 134 lbs.

Explanation :

The balanced chemical reaction will be:

`C_3H_8+5O_2\rightarrow 3CO_2+4H_2O`

Given :

Mass of
`C_3H_8` = 29 lbs = 13154.2 g

conversion used : 1 lbs = 453.592 g

Molar mass of
`C_3H_8` = 44 g/mole

First we have to calculate the moles of
`C_3H_8`.

`\text{ Moles of }C_3H_8=\frac{\text{ Mass of }C_3H_8}{\text{ Molar mass of }C_3H_8}=(13154.2g)/(44g/mole)=298.9moles`

Now we have to calculate the moles of
`CO_2` and
`H_2O`.

From the balanced chemical reaction we conclude that,

As, 1 mole of
`C_3H_8` react to give 3 moles of
`CO_2`

So, 298.9 mole of
`C_3H_8` react to give
`298.9* 3=896.7` moles of
`CO_2`

and,

As, 1 mole of
`C_3H_8` react to give 4 moles of
`H_2O`

So, 298.9 mole of
`C_3H_8` react to give
`298.9* 4=1195.6` moles of
`H_2O`

Now we have to calculate the mass of
`CO_2` and
`H_2O`.

Molar mass of
`CO_2` = 44 g/mole

Molar mass of
`H_2O` = 18 g/mole

`\text{Mass of }CO_2=\text{Moles of }CO_2* \text{Molar mass }CO_2`

`\text{Mass of }CO_2=896.7mole* 44g/mole=39454.8g=86.98lbs`

and,

`\text{Mass of }H_2O=\text{Moles of }H_2O* \text{Molar mass }H_2O`

`\text{Mass of }H_2O=1195.6mole* 18g/mole=21520.8g=47.44lbs`

The total mass of products = Mass of
`CO_2` + Mass of
`H_2O`

The total mass of products = 86.98 + 47.44 = 134.42 ≈ 134 lbs

Therefore, the mass of combustion products formed are 134 lbs.