Question

how many moles of O2 at stp are required to burn completely 4.4 grams of propane
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Answer 1

Answer:

The number of moles of O₂ to burn completely 4.4 grams of propane = 0.5 moles.

Explanation:

The given reaction can be represented as:

C₃H₈ + 5O₂ → 3CO₂+ 4H₂O

From the given balanced chemical equation,

to burn one mole of propane completely we need  5 moles of oxygen.

we know that,

the atomic mass of carbon = 12$g/mol$

the atomic mass of hydrogen = 1 $g/mol$

the atomic mass of oxygen = 16 $g/mol$

from the above data:

molar mass of propane = (12×3) + (1×8)

                                       = 44 $g/mol$

Molar mass of oxygen = 2×16 = 32$g/mol$

From the chemical equation:

1 mole of propane needs = 5 mole of oxygen

that is; 44 grams of propane to burn completely needs = (5×32 ) = 160 grams of oxygen.

For one gram of propane = $\frac{160}{44}$ = 3.64 gram of oxygen

∴ for 4.4 grams of propane = 4.4 × 3.64 = 16 grams of oxygen = 0.5 moles of oxygen.

                                     

Answer 2

For 4.4 grams of propane = 4.4 × 3.64 = 16 grams of oxygen = 0.5 moles of oxygen.

Explanation :-

The given reaction can be represented as:

C₃H₈ + 5O₂ $\longrightarrow$ 3CO₂+ 4H₂O

From the given balanced chemical equation, we observe that

To burn one mole of propane completely we need  5 moles of oxygen.

As we know that,

The atomic mass of Carbon = 12g/mol

The atomic mass of Hydrogen = 1 g/mol

The atomic mass of Oxygen = 16 g/mol

From the above information,

Molar mass of propane = (12 x 3) + (1 x 8)

                                       = 44 g/mol

Molar mass of oxygen = 2 x 16 = 32g/mol

From the chemical equation:

1 mole of propane needs = 5 mole of oxygen

i.e., 44 grams of propane to burn completely needs

= (5 x 32 )

= 160 grams of oxygen

For one gram of propane =  160/44 = 3.64 gram of oxygen

FINAL ANSWER :

∴ For 4.4 grams of propane = 4.4 × 3.64 = 16 grams of oxygen = 0.5 moles of oxygen.