a What is the limiting reactant 5 points b What reactant and how many grams of

A what is the limiting reactant 5 points b what

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aWhat is the limiting reactant? (5 points)bWhat reactant and how many grams of it are left over? (5 points)cHow many grams of carbon dioxide is produced? (5 points)a2CH3OH+ 3O22CO2+ 4H2O2 moles 3 moles 2 molesd = m/V → V = m/dV methanol= 3.20 / 0.79 = 40.5 mL per mole of CH3OH2.00 L of CH3OH = (2000 x 1) / 40.5 = 49.4 molEvery 2 moles of CH3OH will react completely with 3 moles of O249.4 moles of CH3OH will react completely with (49.4 x 3) / 2 = 74.1 moles of O274.1 mol of O2 = (74.1 x 32) g = 2371.2 g = 2400 g of O2CH3OH is the limiting reagentbOxygen has (80,000 – 2400) = 77,600 g left overcEvery 2 moles of CH3OH will produce 2 moles of CO249.4 moles of CH3OH will produce 49.4 moles = (49.4 x 44) g = 2200 g of CO211An iron bar weighed 664 g. After the bar had been standing in moist air for a month, exactlyone-eighth of the iron turned to rust (Fe2O3). Calculate the final mass of the iron bar and rust. (5 points)(Reference: Chang 3.111)The rusting process of the iron bar:4Fe+ 3O2 2 Fe2O3223.2 g 319.2 g1/8 of 664 g of Fe = 83 g of Fe turned to rust → produced (319.2 x 83) / 223.2 = 120 g of Fe2O3Final mass of the iron bar and rust = (664 – 83) + 120 = 701 g

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