Question

Urea, CO(NH2)2, is an inexpensive form of nitrogen fertilizer. It is manufactured on a large scale by reacting ammonia gas and carbon dioxide gas. Water vapour is also made in the process.a) What volume of ammonia gas measured at 37 °C and 105 kPa is required to produce600 g of urea when reacted with carbon dioxide?b) What volume of carbon dioxide gas, is required at the same temperature andpressure?

Answer 1

The Reaction :

2NH3 (g) + CO2 (g) → CO(NH2)2 + H2O

NH3 Ammonia

CO2 Carbon dioxide gas

CO(NH2)2 Urea

H2O water

Don't forget to balance your equation.

a) To get the volume of NH3 we need to assume Ideal gas conditions.

Ideal Gas Law:

`p\text{ x V = n x R x T}`

Step 1) We calculate the moles of NH3 that we need to obtain 600 g of Urea, so:

The molar mass of Urea = 60 g/mol (approx.)

From the reaction, we know that:

2 x 1 mole of NH3 ---------- 60 g of Urea

X ---------- 600 g of Urea

`X=\text{ }\frac{600gUreax2x1moleNH_3}{60\text{ g Urea}}=\text{ 20 moles}`

n (moles) NH3 = 20 moles

Step 2) We clear the volume (V) from Ideal Gas Law:

`\begin{gathered} p\text{ x V = n x R x T} \\ V\text{ = }\frac{n\text{ x R x T}}{p} \end{gathered}`

n = moles of NH3 = 20 moles

T = absolute temperature = 37 °C + 273 = 310 K

p = pressure = 105 kPa

R = gas constant = 8.31 kPa x m3 / mol x K

`V\text{ = }\frac{20\text{ moles x 8.31 }\frac{kPa.m^3}{\text{mol}\mathrm{}K}x310\text{ K}}{105\text{ kPa}}=409.68m^3=410m^3`

Answer: V (NH3) = 410 m3

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b) It is the same procedure above.

Moles of CO2:

1 mol of CO2 ------ 60 g of Urea

Y ------- 600 g of Urea

`Y\text{ = }\frac{600\text{ g x 1 mol}}{60}=10\text{ moles}`

From Ideal gas law:

`V\text{ = }\frac{10\text{ x 8.31 x 310}}{105}=245m^3`

Answer: V(CO2) = 245 m3