Question

For the following reaction, 9.64 grams of carbon (graphite) are allowed to react with 12.5 grams of oxygen gas. carbon (graphite) (6) + oxygen (g) carbon dioxide (g) What is the maximum amount of carbon dioxide that can be formed? grams Visited What is the FORMULA for the limiting reagent? What amount of the excess reagent remains after the reaction is complete? grams

Answer 1

Hhower Carbon (graphite) (s) + Oxyses GJ - Carbon dioxide ) C & t 02 (2 co2 (2) First calcolate mole af & and Ora o mole of C 12.0almol - Weight of C molan mars afc = .64g - 0.803 mol 0.350 mol ③ mole of Oz - weight of O2 = 12.5g Molar mass of 2 32.0 mot From above balence equation, 1 mol c & react with 1 mol O₂ ( 1 mol CO2 o. to produced mol O2 which is in 0.803 mol Co would react with 0.803 but the mole of @ O₂ (g) is 0.390 mol Limited. so limiting reagent - Ozol excess Reagent – co . in 320 g O2 Produced 1.0 g 02 would produced = 44.0g co2 = 44.0 a con CO2 32.0 . 12.5 g O would produced = 440 x12.5 g CO2 32.0 - 550 g co2 320 - 17. 187 g co2 – 17.187g. amount of co2

& excess mol Left After Reaction is = 0.803 – 0.390 = 0.413 mil Cs. mass of Cos) Ces) = = mol of C (s) & 0.913 x 12.0 Molas mass of g - 4.956g excess Reagent After the reaction = The 4.956g Col. "Please rate it"