Moodle@ Moc ×·Course Gener, xV ootal Resan x Seurwet5 16233 11/09/17 83% O This

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Moodle@ Moc ×·Course Gener, xV ootal Resan x Seurwet5 16233 11/09/17 83% O This is a Numeric Entry question/ It is worth 1 point/You have unlimited attempts/There is no attempt penalty 04 Question (1 point) See page 302 One of the intermediates in the synthesis of glycine from ammonia, carbon diaide, and methane is aminoscetonitrile equation is: CHaNG. The balanced chemical 1st attempt see Periodic Table See Hint How much C2H4N2 could be cpected from the reaction of 149 g Co,221 g NH) and 1.79 g CMa? 04 4

Explanation / Answer

For the given intermediate synthesis,

grams of CO2 = 14.9 g

moles of CO2 = 14.9 g/44.1 g/mol = 0.34 mol

grams of NH3 = 2.21 g

moles of NH3 = 2.21 g/17 g/mol = 0.13 mol

grams of CH4 = 1.79 g

moles of CH4 = 1.79 g/16 g/mol = 0.112 mol

If all of CH4 is consumed, we would get = 0.112 x 4/3 = 0.15 mol intermediate

If all of CO2 is consumed, we would get = 0.34 x 4/5 = 0.27 mol intermediate

If all of NH3 is consumed, we would get = 0.13 x 4/8 = 0.065 mol intermediate

Since we get minimum moles of interemdiate formed starting with NH3, this is the limiting reactant.

Thus, grams of C2H4N2 formed = 0.065 mol x 56 g/mol = 3.64 g

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