Question

Use the References to access important values if needed for this question. For the following reaction, 0.452 moles of butane (C4H10) are mixed with 0.459 moles of oxygen gas. butane (C4H10)(g) + oxygen(g) → carbon dioxide(g) + water(g) What is the formula for the limiting reagent? What is the maximum amount of carbon dioxide that can be produced? O moles moles Submit Answer Retry Entire Group 2 more group attempts remaining

Answer 1

First write the balanced equation 2 C4 Higo + 130₂ g. 8CO₂ igo + 10 H219) Now from stoichiometry Every mole of Ca Ho requires : 13 mole Oz = 6.5 mole O₂ for 0. 452 mole C4H1o we need = 0.452 x 6.5 mole O₂ = 2.938 mole 0₂ However, we only have 0.459 mole . Therefore, O₂ is the limiting Reagent- formula for the limiting reagent is O₂ Maximum amount of co, that can be produced - 8 mole co x 0.459 mol O2 13 mole a - 0.28246 mole CO₂ ~ 0.282 mole co.