Question

A 20 cmº sample of butyne, C.H., is bumt in 150 cm of oxygen. This is an excess of
oxygen
2C H.() + 11029) 8C02(9) + 6H,00)
© What volume of oxygen reacts?
11)
What volume of carbon dioxide is produced?
(iii) What is the total volume of gases left at the end of the reaction?
[1]​

Answer 1

• The reaction is CH4+2O2→CO2+2H2O.

At room temperature, water will condense to form liquid water. 1 mole of methane reacts with 2 moles of oxygen to form 1 mole of oxygen.

10 ml (0.445 moles) of methane will react with 20 ml (0.89 moles) of oxygen to form 10 ml (0.445 moles) of carbon dioxide.Hence, the resulting mixture will contain 10 ml of unreacted methane and 10 ml of carbon dioxide.

The total volume of the mixture will be 20 ml.

• To be able to compare our experimental value to the standard molar volume (22.4 L/mol), we must determine the molar volume at STP (0° C, 1atm). We can use the combined gas law to find the volume the sample of oxygen gas created would have at STP.

• The volume of one mole of CO2 produced is 24 dm³ at room temperature and pressure. Alternatively, if your reaction took place at standard temperature and pressure (273 K, 1 atm), then the molar volume is 22.4 dm³.

• Here's a breakdown: First, you identify that ethyne is the limiting reagent and would be consumed completely. Second, you correctly calculated the volume of oxygen left that is 450 cm 3450cm³. 20cm 320 cm³ of ethyne will completely react to produce 40 cm340 cm3 of carbon dioxide.

Thus, the total gas volume when the reaction reaches completion is 490cm³ 490cm³.