N2+3H2 yields 2NH3 at STP How many grams of nitrogen are needed to react with 1.

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1) Grams of N2 per 10.7g of H2 a) Balance the equation: N2 + 3H2 ----> 2NH3 to define how many moles of N2 / mole H2 (1:3) b) Calculate moles of H2: grams/ molecular weight = moles ---> 10.7 / 2 = 5.35 moles c) Moles of N2 per mole of H2: 5.35 moles of H2 x 1 mole N2 / 3 moles H2 = 1.783 moles N2 d) Grams of N2: Moles N2 x Grams per mole aka molecular weight of N2 = Weight ---> 1.783 moles N2 x 28 (grams/mole) = 49.93; ANSWER: *** say 50 grams N2 ***needed to completely react with the 10.7 g of H2 2) Liters of H2 needed to react with xs CO2 and produce 28.0g CH3OH a) Balance the equation: 2H2 + CO ---> CH3OH b) Calculate moles of CH3OH 28 g / 32g/mole = 0.875 moles c) Calculate moles of H2 needed to react with excess CO and produce 0.875 moles CH3OH: 0.875 moles CH3OH x 2 moles H2 per mole of CH3OH (per the balanced equation) = 1.75 moles H2 d) Calculate volume of H2 used: assume STP to calculate volume of H2 (since no other data was provided) so V = nRT / P = (1.75 moles)(0.08205 L-Atm / mole-K.) (273 K.) / (1 atm) = 39.2 Liters ANSWER *** 39.2 Liters H2 *** to produce 28g CH3OH in the presence of excess CO

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