Question

I am having trouble with these questions. Can anybody help me please.

Given the following correctly BALANCED reaction: PENTANE (72 g/mol) 5. (3 pts) How grams of CO2 are created from 6. (3 pts) How many grams of pentane will be needed t 1440 grams of CsH12 (pentane). c5H12 + 802 → 5CO2 + 6H20 ΔΗ=-840 kcals/mol create 250 kilocalories? (the energy in a candy bar) 7. (2 pts) When 100 moles of pentane (CsH2) are mixed with 20 kilograms of oxygen, which substance is the limiting reagent? 8. (1 pts) How many moles of the excess reagent remain after complete reaction?

Answer 1

CsH,, + 802 → 5CO2 + 6H20 AH=-840 kcals/mol 5) No. of moles of pentane (1440 g) (72 g/mol) 20 mol No. of moles of CO2 produced (5/ 1) x 20 100 mol weight of CO2 produced (100 mol) x (44.01 g/mol) 4401 gramd of C02 6) 1 mol of pentane produced heat 840 No. of moles of pentane needed to produce 250 kCal - (1 mol / 840) x 250 kCal -0.2976 mol Weight of pentane needed - (0.2976 mol) x (72 g/mol) 21.4 grams of pentane 7) No. of moles of Oxygen (20000 g) / (32 g/mol) 625 mol No. of moles of O2 needed to react with 100 mol of pentane (811) x 100 mol 800 mol Here, Oxygen present in limited quantity So, the limiting reagent is Oxygen 8) Moles of petane reacted with 625 mol of oxygen (1/ 8) x 625 mol- 78.125 mol excess pentane remain 100 mol 78.125 mol 21,875 mol of pentane remain