Question

how many litres of oxygen at stp are required to burn completely 2.2g of propane ,c3h8? and also tell at 11.2L how many oxygen in g will go through complete combustion ​

Answer 1

Answer:

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Explanation:

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Answer 2

Given:

Propane = 2.2g

Oxygen = 11.2L

To find:

i)The liter of oxygen at STP is required to burn completely 2.2g of propane.

ii) The oxygen in the gram will go through complete combustion.

Solution:

i) C3H8 + 5O2 -> 3CO2 + H2O

   44g       5 x 22.4 L

(1mole)       (at STP)

1 mole(44g) = 5 X 22.4 O2 at STP

2.2 g of propane = 5 x 22.4 x 2.2/ 44

                            = 5.6L of O2 at STP

Hence, the liter of oxygen at STP required to burn completely 2.2g of propane is 5.6L.

​ii) C3H8 + 5O2 -> 3CO2 + H2O

   160 g at 22.4 L at STP

   Oxygen in grams for 11.2 L= 11.2 X 160/22.4

                                               = 80g

Hence, 80 gram of the oxygen will go through complete combustion.