Question

For the combustion of propane (c3h8), how many grams of oxygen are needed to fully react 45.0 grams of propane? report your answer to the nearest whole number.

Answer 1

For the combustion of propane $C_{3}H_{8}$, 164 grams of oxygen is needed to fully react 45.0 grams of propane.

We know that,

The combustion of propane is as follows:-

$C_{3}H_{8}+5O_{2}$ → $3CO_{2}+4H_{2}O$

Hence, for 1 mole of propane 5 moles of $O_{2}$ are needed.

We know that,

Mass of 1 mole of propane = 12*3 + 1*8 = 36+8 = 44g

Hence,

45g of propane = 45/44 moles of propane = 1.022 moles.

Hence,

1.022 moles of propane will need 5*1.022 = 5.11 moles of $O_{2}$.

1 mole of $O_{2}$ = 2*16 = 32g

5.11 moles of $O_{2}$ = 32*5.11 = 163.52g ≈ 164g $C_{3}H_{8}$, 164 grams of oxygen are needed to fully react 45.0 grams of propane.

We know that,

The combustion of propane is as follows:-

$C_{3}H_{8}+5O_{2}$ → $3CO_{2}+4H_{2}O$

Hence, for 1 mole of propane 5 moles of $O_{2}$ are needed.

We know that,

Mass of 1 mole of propane = 12*3 + 1*8 = 36+8 = 44g

Hence,

45g of propane = 45/44 moles of propane = 1.022 moles.

Hence,

1.022 moles of propane will need 5*1.022 = 5.11 moles of $O_{2}$.

1 mole of $O_{2}$ = 2*16 = 32g

5.11 moles of $O_{2}$ = 32*5.11 = 163.52g ≈ 164 grams.

To learn more about combustion of propane, here:-

https://brainly.com/question/12328568

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