Question

One of the intermediates in the synthesis of glycine from ammonia, carbon dioxide, and methane is aminoacetonitrile, C2H4N2.

How much C2H4N2 could be expected from the reaction of 14.9 g CO2, 2.11 g NH3, and 1.75 g CH4?

Answer 1

• You could expect 3.48 grams of C₂H₄N₂

Step-by-step explanation:

You must start by stating the chemical equation for the reaction of ammonia, carbon dioxide, and methane to produce aminoaceto nitrile.

1. Word equation:

Ammonia + Carbon dioxide + Methane → Aminoacetonitrile + Water

2. Balanced chemical equation:

`8NH_3+5CO_2+3CH_4\rightarrow 4C_2H_4N_2 +10H_2O`

3. Convert the mass of each reactant into number of moles:

Formula:

• Number of moles = mass in grams/molar mass

2.11g NH₃

• Number of moles = 2.11g / 17.03g/mol = 0.124 mol NH₃

14.9g CO₂

• Number of moles = 14.9g/44.01g/mol = 0.339 mol CO₂

1.75g CH₄

• Number of moles = 1.75g/16.04g/mol = 0.109 mol CH₄

4. Theoretical mol ratio

From the balanced chemical equation, using the coefficientes:

`8molNH_3:5molCO_2:3molCH_4:4molC_2H_4N_2:10molH_2O`

5. Limiting reagent

The available amounts of the reactants are:

• 0.124 mol NH₃

• 0.339 mol CO₂

• 0.109 mol CH₄

Fom the theoretical mole ration, to react with 0.124 mol of NH₃ you would need:

• 0.124molNH₃ × (5molCO₂/8molNH₃) = 0.0775 mol CO₂

Since there are 0.339 moles available, this is in excess.

• 0.124molNH₃ × (3molCH₄/8molNH₃) = 0.0465mol CO₂

Since there are 0.109 moles available, this is in excess too.

Hence, the limiting reagent is NH₃.

6. Yield

Use the theoretical ratio:

• 0.124molNH₃ × (4molC₂H₄N₂ / 8molNH₃) = 0.0620 mol C₂H₄N₂

Convert to grams:

• Mass = number of moles × molar mass

• 0..0620 mol × 56.068g/mol = 3.48 g of C₂H₄N₂ ← answer