Question

One of the intermediates in the synthesis of glycine from ammonia , carbondioxide and methane is aminoacetonitrile C2H4N2. The balanced chemical equation is 3CH4+5 CO2 + 8NH3 ---> 4C2H4N2 +10H2O. How much C2H4N2 could be expected from the reaction of 13.2 g CO2, 2.18 g NH3 and 17.0 g CH4

Answer 1

molar mass CH4 = 16

molar mass CO2 = 44

molar mass NH3 = 17

mass of CO2 = 13.2 g

mass of NH3 = 2.18 g

mass of CH4 = 17 g

moles = mass / molar mass

moles of co2 = 13.2 /44 =0.3 moles

moles of Ch4 = 17 / 16 = 1.0625 moles

moles of NH3 = 2.18 / 17 = 0.128 moles

3 CH₄+5 CO₂ + 8 NH₃ ---> 4 C₂H₄N₂ +10 H₂O

calculate the moles needed per reactant to find the limiting reactant:

0.3 moles of CO2 will require:

moles of CH4; 5 moles of CO2 will react with 3 moles of methane

0.3 moles of CO2 x 3 moles methane/5 moles CO2 = 0.18 moles of CH4

moles of NH3; 5 moles of CO2 will react with 8 moles of NH3

0.3 moles of CO2 x 8 moles NH3 / 5 moles CO2 = 0.48 moles of NH3

In this case The ammonia is not enough to react with all the co2,

Make calculations for ammonia:

moles of ammonia = 0.128 moles

calculate moles of methane, 3 moles of methane require 8 moles of ammonia

0.128 moles of ammonia x 3 moles of methane / 8 moles of ammonia

moles of methane = 0.128 * 3 / 8 = 0.048 moles of methane

since there is enough methane then we can say that ammonia is the limiting reactant so stoichiometrical calculations must be done considering the compound:

moles of ammonia = 0.128 moles

moles of CO2 = 0.128 moles of nh3 * 5 moles of co2 / 8 moles ammonia = 0.128*5/8 = 0.08 moles of CO2

moles of methane = 0.128 moles of nh3 * 3 moles of CH4 / 8 moles of NH3 = 0.048 moles of methane

now calculate the ammount of aminoacetonitrile:

8 moles of ammonia produce 4 moles of acetonitrile:

moles of acetonitrile = 0.128 * 4 / 8 = 0.064 moles of aminoacetonitrile

molar mass of aminoacetonitrile is 56.066 g/mol

mass = moles * molar mass = 0.064 * 56.066 = 3.59 grams of aminoacetonitrile

*hope this helps =)