Respuesta :
Answer : The mass of combustion products formed are 134 lbs.
Explanation :
The balanced chemical reaction will be:
[tex]C_3H_8+5O_2\rightarrow 3CO_2+4H_2O[/tex]
Given :
Mass of [tex]C_3H_8[/tex] = 29 lbs = 13154.2 g
conversion used : 1 lbs = 453.592 g
Molar mass of [tex]C_3H_8[/tex] = 44 g/mole
First we have to calculate the moles of [tex]C_3H_8[/tex].
[tex]\text{ Moles of }C_3H_8=\frac{\text{ Mass of }C_3H_8}{\text{ Molar mass of }C_3H_8}=\frac{13154.2g}{44g/mole}=298.9moles[/tex]
Now we have to calculate the moles of [tex]CO_2[/tex] and [tex]H_2O[/tex].
From the balanced chemical reaction we conclude that,
As, 1 mole of [tex]C_3H_8[/tex] react to give 3 moles of [tex]CO_2[/tex]
So, 298.9 mole of [tex]C_3H_8[/tex] react to give [tex]298.9\times 3=896.7[/tex] moles of [tex]CO_2[/tex]
and,
As, 1 mole of [tex]C_3H_8[/tex] react to give 4 moles of [tex]H_2O[/tex]
So, 298.9 mole of [tex]C_3H_8[/tex] react to give [tex]298.9\times 4=1195.6[/tex] moles of [tex]H_2O[/tex]
Now we have to calculate the mass of [tex]CO_2[/tex] and [tex]H_2O[/tex].
Molar mass of [tex]CO_2[/tex] = 44 g/mole
Molar mass of [tex]H_2O[/tex] = 18 g/mole
[tex]\text{Mass of }CO_2=\text{Moles of }CO_2\times \text{Molar mass }CO_2[/tex]
[tex]\text{Mass of }CO_2=896.7mole\times 44g/mole=39454.8g=86.98lbs[/tex]
and,
[tex]\text{Mass of }H_2O=\text{Moles of }H_2O\times \text{Molar mass }H_2O[/tex]
[tex]\text{Mass of }H_2O=1195.6mole\times 18g/mole=21520.8g=47.44lbs[/tex]
The total mass of products = Mass of [tex]CO_2[/tex] + Mass of [tex]H_2O[/tex]
The total mass of products = 86.98 + 47.44 = 134.42 ≈ 134 lbs
Therefore, the mass of combustion products formed are 134 lbs.
