Respuesta :
The 35.1 kg of carbon dioxide produced on combustion of 18.9 Liters of propane.
Given:
18.9 Liters of liquid propane with a density of 0.621 g/mL
To find:
The mass of carbon dioxide gas produced on combustion of 18.9 L of propane.
Solution:
The mass of propane = m
The volume of the propane = v = 18.9 L
[tex]1 L = 1000 mL\\18.9 l=18.9\time 1000 mL=18,900 mL[/tex]
The density of the liquid propane = d = 0.621 g/mL
[tex]d=\frac{m}{v}\\m=d\times v\\m=0.621 g/mL\times 18,900 mL=11,736.9 g[/tex]
The mass of propane combusted = 11,736.9 g
The chemical reaction of combustion of propane
[tex]C_3H_8(l)+5O_2(g)\rightarrow 3CO_2(g)+4H_2O(g)[/tex]
The molar mass of propane = 44.1 g/mol
Moles of propane :
[tex]\frac{11,736.9 g}{44.1 g/mol}=266.143 mol[/tex]
According to the reaction, 1 mole of propane gives 3 moles of carbon dioxide gas, then 266.143 moles of propane will give:
[tex]=\frac{3}{1}\times 266.143 mol=798.429 \text{mol of}CO_2[/tex]
Mass of 798.429 moles of carbon dioxide gas:
[tex]=798.429 mol\times 44.01 g/mol=35,138.86 g\\\\1 g = 0.001 kg\\\\35,138.86 g=35,138.86\times 0.001k g\\\\=35.13886 kg\approx 35.1 kg[/tex]
The 35.1 kg of carbon dioxide produced on combustion of 18.9 Liters of propane.
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