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The balanced reaction is:

C3H8 + 5O2 = 3CO2 + 4H2O

We are given the amount of C3H8 to be reacted with O2. This will be the starting point of the calculations.

73.7 g C3H8 ( 1 mol C3H8/ 44.1 g C3H8 ) ( 5 mol O2 / 1 mol C3H8 ) ( 32.0 g O2 / 1 mol O2 ) = 267.39 g O2

Answer: 267.2 grams

Explanation:

[tex]C_3H_8+5O_2\rightarrow 3CO_2+4H_2O[/tex]

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]

[tex]\text{Number of moles of propane}=\frac{73.7g}{44.1g/mol}=1.67moles[/tex]

According to stoichiometry:

1 mole of [tex]C_3H_8[/tex] require 5 moles of [tex]O_2[/tex]

Thus 1.67 moles of [tex]C_3H_8[/tex] will require =[tex]\frac{5}{1}\times 1.67=8.35moles[/tex] of [tex]O_2[/tex]

Mass of [tex]O_2=moles\times {\text {Molar mass}}=8.35moles\times 32g/mol=267.2[/tex]

Thus 267.2 grams of oxygen are required.

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