Week 8/9 HW 04/08/18 76% mail stsu.e This is a Numeric Entry question/It is wort

Question

Week 8/9 HW 04/08/18 76% mail stsu.e This is a Numeric Entry question/It is worth 1 point/You have unlimited attempts/There is no attempt penalty 08 Question (1 point) a See page 302 The industrial synthesis of H2 begins with the steam- reforming reaction, in which methane reacts with high-temperature steam: CH4(g) + H2O(g)- CO2(g) + 3H2(g) 4th attempt al See Periodic Table See Hint What is the percent yield when a reaction vessel that initially contains 67.5 kg CH4 and excess steam yields 14.0 kg H2? 58.3 > 3rd attempt > 2nd attempt

Explanation / Answer

CH4(g)   + H2O(g) ---------- CO2(g)   + 3H2(g)

mass of CH4 = 67.5 Kg

molar mass of CH4 = 16.04 gram/mole

molar mass of CH4= 16.04/1000 = 0.01604 kg/mole

number of moles of CH4= mass/molarmass = 67.5kg/0.01604kg/mole

number of moles of CH4 = 4208.23 moles

limiting reagent is CH4 because steam is excess.

according to equation

1 mole of CH4 = 3 moles of H2

4208.23 moles of CH4 = ?

                                 =4208.23x3/1= 12624.69 moles of H2

number of moles of H2 formed = 12624.69 moles

molar mass of H2 = 2.016 gram/mole

molar mass of H2 = 2.016/1000 kg/mole = 0.002016 kg/mole

mass of one mole of H2 = 0.002016 kg/mole

mass of 12624.69 moles of H2 =?

                                           = 12624.69x0.002016/1 =25.45 Kg

theoritical yield of H2 = 25.45 Kg

Actual yield = 14.0 Kg

Percent yield = actual yield/theortical yield x100

Percent yield = 14.0/25.45 x100 = 55.00%

percent yield= 55.0%

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