2C6H6(l) +1 502(g)-12CO2(g) + 6H2O(g)2 (g)(A) 74 I(B) 11.2 1(C) 22.4 L(D) 84 L

2C6H6(l) +1 502(g)-12CO2(g) + 6H2O(g)2 (g)(A) 74 I(B) 11.2 1(C) 22.4 L

1.	2C6H6(l) +1 502(g)-12CO2(g) + 6H2O(g)2 (g)(A) 74 I(B) 11.2 1(C) 22.4 L(D) 84 L
Answer» <p>Step 1: we need to find out 39g of liquid benzene is equivalent to how many moles of the liquid.</p><p>Moles = mass/molar massmass = 39gmolar mass= 78</p><p>Therefore moles of liquid benzene = 39/78 = 0.5 moles</p><p>Step 2: Use the equation plus the moles of the liquid benzene to find the moles of O₂ used.</p><p>2C₆H₆+15O₂--->12CO₂+ 6H₂O</p><p>We see that for every 2 moles of liquid benzene we require 15 moles of O₂ for the reaction. Thus;</p><p>If 2 moles benzene needs 15 moles of oxygen,Then 0.5 moles will need 15ₓ 0.5/2 = 3.75 moles of oxygen.</p><p>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)</p><p>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</p><p>Therefore 84 liters of oxygen is what is required to completely burn 39g of liquid benzene.</p>