Question

For the following reaction, 7.82 grams of carbon (graphite) are allowed to react with 11.0 grams of oxygen gas. carbon (graphite) (s) + oxygen (g) — carbon dioxide (g) What is the maximum amount of carbon dioxide that can be formed? grams What is the FORMULA for the limiting reagent? What amount of the excess reagent remains after the reaction is complete? grams Submit Answer Retry Entire Group No more group attempts remain

Answer 1

C(s) + O2(g) → CO2 (g) Given, mass of Carbon = 7.82g mass of oxygen gas = 110g. Molar mass of Carbon = 12g/mol molar mass of oxygen gas (02) = 32g/mol mote - mals (given) Motar mass So, molesd Carbon = 182g = 0.6smor 72 g/mot motes of oxygen = 11.0g = 0.34 mor 32 87 mot From the above reaction we find that I more of C reacts with Imore of Og. So, 0.65 mol of C will react with 0.65 of O, but only 0.34 mord og is available. . so, o is limiting reagent since, Imole of c forms more of Cogo in presence of imore &r. T so, Excers amount of carbon = (0.65mor-0.34 moi) = 0.31 mor Here 0.34 mor of Carbon and 0.34 more of oxygen will form 0.34 more of Carbon dioxide (CO₂) - Molar mass of Co₂ = 44 g/mol mosso comotar mass * more = 0.34 moix 44 g/mol = 14.96g Hence, 14.96g of Co2 will be produced.