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8.Liquid benzene (CoHs) burns in oxygen accordingto 2C,H, ()+1502(9) 12CO2(g)+ 6H2O(g)How many litres of o2 at STP are needed tocomplete the combustion of 39 g of liquid benzene?(1) 74 L(3) 22.4 L(2) 11.2 L4 84 L |
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Answer» Step 1: we need to find out 39g of liquid benzene is equivalent to how many moles of the liquid. Moles = mass/molar massmass = 39gmolar mass= 78 Therefore moles of liquid benzene = 39/78 = 0.5 moles Step 2: Use the equation plus the moles of the liquid benzene to find the moles of O₂ used. 2C₆H₆+15O₂--->12CO₂+ 6H₂O We see that for every 2 moles of liquid benzene we require 15 moles of O₂ for the reaction. Thus; If 2 moles benzene needs 15 moles of oxygen,Then 0.5 moles will need 15ₓ 0.5/2 = 3.75 moles of oxygen. Step 3: Calculate the volume of oxygen required from its moles calculatedThe ideal gas law states that 1 mole of an ideal gas will occupy 22.4 liters at STP(standard temperature and pressure) Therefore if 1 mole of oxygen occupies 22.4 liters,Then 3.75 moles of oxygen will occupy; 22.4 litersₓ 3.75 moles/1 mole = 84 liters Therefore 84 liters of oxygen is what is required to completely burn 39g of liquid benzene. |
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