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The combustion of propane may be described by the chemical equation C 3 H 8 ( g ) + 5 O 2 ( g ) ⟶ 3 CO 2 ( g ) + 4 H 2 O ( g ) How many grams of O 2 ( g ) are needed to completely burn 91.6 g C 3 H 8 ( g ) ?

Answer :

Answer : The mass of [tex]O_2[/tex]  needed are, 332.8 grams

Solution : Given,

Mass of [tex]C_3H_8[/tex] = 91.6 g

Molar mass of [tex]C_3H_8[/tex] = 44 g/mole

Molar mass of [tex]O_2[/tex] = 32 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{91.6g}{44g/mole}=2.08moles[/tex]

Now we have to calculate the moles of [tex]O_2[/tex]

The balanced chemical reaction is,

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

From the reaction, we conclude that

As, 1 mole of [tex]C_3H_8[/tex] react with 5 mole of [tex]O_2[/tex]

So, 2.08 moles of [tex]C_3H_8[/tex] react with [tex]2.08\times 5=10.4[/tex] moles of [tex]O_2[/tex]

Now we have to calculate the mass of [tex]O_2[/tex]

[tex]\text{ Mass of }O_2=\text{ Moles of }O_2\times \text{ Molar mass of }O_2[/tex]

[tex]\text{ Mass of }O_2=(10.4moles)\times (32g/mole)=332.8g[/tex]

Thus, the mass of [tex]O_2[/tex]  needed are, 332.8 grams

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