Question

Name TA Laboratory Date - Section Experiment 9-Post-lab Questions 1. Methanol (CH,OH) can be produced by the reaction of carbon monoxide (CO) with hydrogen (H). The balanced chemical reaction is: CÓg) + 2 H (8) -CH,OH() Suppose 14.0 g of carbon monoxide is reacted with 8.0 g of hydrogen. a. Which reactant is the limiting reagent? b. What is the theoretical yield for methanol? c. At the completion of the reaction, how many grams of the excess reagent will remain? 2. Iron(III) oxide (Fe,,) can be converted to iron metal (Fe) by the reaction with carbon (C) where the other product is carbon dioxide (CO). The balanced chemical reaction is: 2 Fe, 0, (s) + 3 C(s) + 4 Fe(s) + 3 CO, (c) Suppose 3 moles of iron(III) oxide is reacted, a. How many moles of carbon are needed to react exactly with the iron oxide with no amount of either reactant left over? b. How many moles of iron will be produced? c. How many moles of carbon dioxide will be produced?

Answer 1

Q1. Colg) + 2H₂ (g) - CH₂OHU. Moles of carbon moncoxide Mass Molae mass. 149 0.5 meles. - 289/mel 28g1 mee meres of H₂ = 8.0g - 4 moles 2g/mol. (a) - - (6) If all of the 4 meres of the were to be used up there would need to be 4 moles H2 xlmolco 2meles H₂ = 2meles of H₂. But there is only 0.5 meles of co, Thulefore co is the limiting reagent Amount of limiting reagent will decide the amount (theoretical field) of product formation from above balance Equation: 1 mole co broduce one mole ol CH₂OH. 0.5 meles co will produce = 1 mol CH3OH x D Smalco Imel Co = 0.5 mel CH₂OH. Theoretical yield of CHOL - Moles se Molae mass - 0.5 mol x 32g/mol. Theoretical field = 16g.

(o to reacts H₂ is present in excess. from the balanced Equation, I mel with 2 mol H₂. 10.5 mel co will reacts with a 2 mol H₂ X D. Smel co = 1 mol H₂ 1 met co Moles of excess H₂ = 4moles - 1 mol = 3 meles. Mass of excess H₂ - Moles & Molar mass. - 3 ml x 29/met. mass of excess reagent = 6g

2 2 Fe₂O₂ (s) + 3cu 4 Fe(s) + 300g 2 meles Fe2O₂ reacts with 3 mel c. 3 meles fe, O₂ will reacts with 3 melc x 3 mal Fee O₂- 2 med Fe₂O₃ = 4.5 mac (6) 2 mol Fe₂O₂ will produce 4 mol fe. 3 mel Fe₂O₃ will 4 no 2nd produces : fe x 3 md Fe₂O3 Fe₂O3 = 16 mel fez O3 (O 2 mel 3 mel Fe₂O produces 3 mel co₂. fezl, will produce 3 mel CO2 x 3 me fez Oz 2nd Fe₂O3 = (4.5 mol CO2 Answer will be use amount reactant reacts same for (b) & (c) if we of c. Because both completely.

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