Question

# How much C2H4N2 could be expected from the reaction of13.5 g CO2,2.23 g NH3, and1.63 g...

How much C2H4N2 could be expected from the reaction of13.5 g CO2,2.23 g NH3, and1.63 g CH4?

The reaction:

CO2 + NH3 + CH4 = C2H4N2 + H2O

balance

3CH4 + 5CO2 + 8NH3 = 4C2H4N2 + 10H2O

now...

mol of CH4 = mass/MW = 1.63/16 = 0.1018

mol of CO2 = mass/MW = 13.5/44= 0.3068

mol of NH3 = masS/MW = 2.23/17= 0.1311

ratios are:

3CH4 + 5CO2 + 8NH3

clearly, NH3 is limiting

0.1311 mol requires -->

3/8*0.1311 = 0.0491 mol of CH4

5/8*0.1311 = 0.0819 mol of CO2

then

ratio is

8NH3 = 4C2H4N2

or 2:1

0.1311 mol o fNH3 --> 4/8*0.1311 = 0.06555 mol of C2H4N2

mass = mol*MW = 0.06555 *56.0666 = 3.6751 g C2H4N2

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