Ammonium sulfate, an important chemical fertilizer, can be prepared by the reaction of ammonia with sulfuric acid according to the following balanced equation: 2 NH3(g) + H2SO4(aq) → (NH4)2SO4(aq)

If a reaction vessel has 3.29 L of NH3 at 31.5°C and 22.7 atm, how many grams of H2SO4 are needed to completely react with it?

g H2SO4. Do NOT enter unit. Report your final answer with 3 SFs.

147 grams

Explanation:

We are given:

Pressure (P) = 22.7 atm

Temperature (T) = 304.5 K

Volume (V) = 3.29 L

Solving for the number of moles of Ammonia:

From the Ideal Gas Equation:

PV = nRT

replacing the variables

(22.7)(3.29) = n (0.082)(304.5)           [R = 0.082 L atm / mol K]

n = 3 moles (approx)

Number of moles of Sulphuric Acid required:

We are given the balanced equation:

2 NH3(g) + H2SO4(aq) → (NH4)2SO4(aq)

we can see that for 2 moles of Ammonia, we will need 1 mole of Sulphuric acid

So, we can say that we need half the number of moles of sulphuric acid as compared to Ammonia

Hence, We will need half the number of moles of Ammonia

Number of moles of Sulphuric acid required = 1/2 * Moles of Ammonia

Moles of Sulphuric Acid = 1/2 * 3

Moles of Sulphuric acid = 3/2 moles

Mass of Sulphuric Acid needed:

We know that the molar mass of Sulphuric acid is 98 grams/mol

We know that the mass of a given number of moles of a compound is the number of moles multiplied by the molar mass

Mass of Sulphuric Acid = Number of moles * Molar mass

Mass of Sulphuric Acid = 1.5 * 98

Mass of Sulphuric Acid = 147 grams