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Balance the following equation: __C3H8 __O2 __CO2 __H2O
propane + oxygen  carbon dioxide + water
and answer questions #15 & #16 based on the balanced equation.
15. How many dm3 of oxygen at STP would be required to react completely with 38.8g of propane? (3 marks)
** 1dm3 = 1000mL
16. How many grams of water can be generated from the combustion of 40.0g of propane in an excess of oxygen? (3 marks)

Respuesta :

The number of moles of propane is used to obtain the required results in each case. It is gotten from the equation of the reaction.

a) The balanced equation of the reaction is;

C3H8 + 5O2 -----> 3CO2 + 4H2O

Number of moles of propane = 38.8g/44 g/mol = 0.88 moles

From the reaction equation;

1 mole of propane reacts with 5 mole of oxygen

0.88 moles of propane reacts with 0.88 × 5/1 = 4.4 moles of oxygen.

If 1 mole of oxygen occupies 22.4 dm^3

4.4 moles of oxygen occupies 4.4 moles × 22.4 dm^3/ 1 mole= 98.56 dm^3 of oxygen

b) Number of moles of propane =40.0g /44 g/mol = 0.91 moles of propane

1 mole of propane yields 4 moles of water

0.91 moles of propane yields 0.91 × 4/1 = 3.64 moles of water

Mass of water = 3.64 moles of water × 18 g/mol = 65.52 g of water

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