Please use dimensional analysis to answer these 2. Benzene C6H6 burns in oxygen to produce carbon dioxide and water. D. How many moles of oxygen gas are required to make 40.00 moles of carbon dioxide?O=12+,3=ISc. How many moles of C6H6 must be used to produce 180.0 grams of water?d. How many liters of oxygen at STP are needed to react with 6.00 X 1024 molecules of C6H6?

The balanced equation for the reaction of benzene (C6H6) with O2(g) to form carbon dioxide, CO2(g) and water, H2O(g) is: 2C6H6(l) + 15O2(g) -----> 12CO2(g) + 6H2O(g) #(b): Moles of carbon dioxide, CO2(g) required = 40.00 mol Mole ratio between oxygen, O2(g) and carbon dioxide, CO2(g) is: 15 mol O212 mol CO2 Moles of oxygen, O2(g) required = 40.00 mol CO2 * 15 mol O212 mol CO2= 50.00 mol O2 Moles of oxygen, O2(g) required = 50.00 mol (Answer) #(c): mass of water, H2O(g) requied = 180.0 g Molar mass of H2O(g) = 18.015 g/mol moles of water, H2O(g) requied = 180.0 g * 1 mol H2O18.015 g= 9.9916736 mol H2O Mole ratio between benzene (C6H6) and water, H2O(g) is: 2 mol C6H66 mol H2O Moles of benzene required = 9.9916736 mol H2O * 2 mol C6H66 mol H2O= 3.33 mol C6H6 Moles of benzene (C6H6) required = 3.33 mol (Answer) #(d): Given the molecules of benzene (C6H6) = 6.00*1024 1 mol = 6.023*1023 molecule => moles of benzene (C6H6) = 6.00*1024 molecule benzene * 1 mol benzene6.023*1023 molecule benzene= 9.961813 mol benzene Mole ratio between benzene (C6H6) and oxygen, O2(g) is: 15 mol O22 mol C6H6 moles of O2 required = 9.961813 mol C6H6 * 15 mol O22 mol C6H6= 74.7136 mol O2 At STP, 1 mol O2 is approximately 22.4 L Liters of oxygen, O2(g) required = 74.7136 mol O2 * 22.4 L1 mol O2= 1674 L Liters of oxygen, O2(g) required at STP = 1674 L (Answer)