Question

Ammonium sulfate, an important fertilizer, can be prepared by the reaction of ammonia with sulfuric acid according to the following balanced equation:2NH3 (g) + 1 H2SO4 = 1 (NH4)2SO4 (aq)Calculate the volume of NH3, needed at 24 C and 25 atm to react with 1500 g of H2SO4

Answer 1

To find the volume of NH3 needed to react with 1500 g of H2SO4, calculate the moles of H2SO4, use the 2:1 mole ratio from the balanced equation, and apply the ideal gas law at the given temperature and pressure.

Step-by-step explanation:

To calculate the volume of NH3 needed to react with 1500 g of H2SO4, we need to perform a series of steps. First, we find the number of moles of H2SO4 using its molar mass. Then using the balanced chemical equation 2NH3 (g) + 1 H2SO4 (aq) → (NH4)2SO4 (aq), we find the molar ratio of NH3 to H2SO4, which is 2:1. Finally, we use the ideal gas law, PV=nRT, to calculate the volume of NH3 under the given conditions (24°C and 25 atm).

• Calculate the number of moles of H2SO4: n = mass / molar mass.
• Determine the moles of NH3 using the mole ratio from the balanced equation.
• Convert the temperature to Kelvin by adding 273.15 to the Celsius temperature.
• Apply the ideal gas law (PV=nRT) to calculate the volume of NH3.

Answer 2

FIRST PART

Step 1

The reaction must be completed and balanced:

2NH3 (g) + 1 H2SO4 = 1 (NH4)2SO4 (aq)

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Step 2

The moles of NH3 are calculated by stoichiometry:

Information needed:

The molar mass of H2SO4 = 98 g/mol

Information provided:

The mass of H2SO4 = 1500 g

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Step 3

Procedure:

2 moles NH3 ------- 98 g H2SO4

X ------- 1500 g H2SO4

X = 1500 g H2SO4 x 2 moles NH3/ 98 g H2SO4 = 30.6 moles NH3

First part: Moles of NH3 = 30.6 moles

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SECOND PART

Step 4

NH3 gas is assumed to be ideal, therefore, it is applied:

p x V = n x R x T (1)

p = pressure = 25 atm

T = temperature (absolute) = 24 °C + 273 = 297 K

n = moles (from part 1) = 30.6 moles

R = gas constant = 0.082 atm L/mol K

V = volume = unknown

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Step 5

V is cleared from (1):

V = n x R x T/p = (30.6 moles x 0.082 atm L/mol K x 297 K)/25 atm = 29.8 L

Answer: Volume NH3 = 29.8 L