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 4.56 L of NH3 at 32.3°C and 20.8 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.

SO... I did my work on paper and the answer I got was 2900. Please let me know if i got the answer right.

To find the number of grams of H2SO4 needed to completely react with the given amount of NH3, we can use the ideal gas law equation, PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.

First, we need to calculate the number of moles of NH3 present in the given volume. We can use the ideal gas law equation to do this:

n(NH3) = (PV)/(RT)

P = 20.8 atm
V = 4.56 L
R = 0.0821 atm·L/mol·K (ideal gas constant)
T = 32.3°C + 273.15 = 305.45 K (temperature in Kelvin)

Plugging in these values:

n(NH3) = (20.8 atm) * (4.56 L) / (0.0821 atm·L/mol·K * 305.45 K)

n(NH3) ≈ 1.123 moles

Now, we can see that the stoichiometry between NH3 and H2SO4 is 2:1, meaning that for every 2 moles of NH3, we need 1 mole of H2SO4. Therefore, we need half the number of moles of NH3 in H2SO4:

n(H2SO4) = 1.123 moles / 2 = 0.5615 moles

To convert moles of H2SO4 to grams, we need to know the molar mass of H2SO4, which is calculated by summing the atomic masses of its constituents:

H2SO4: (2 * 1.008 g/mol) + (32.07 g/mol) + (4 * 16.00 g/mol) ≈ 98.09 g/mol

Finally, we can calculate the mass of H2SO4:

mass(H2SO4) = n(H2SO4) * molar mass(H2SO4)

mass(H2SO4) = 0.5615 mol * 98.09 g/mol ≈ 55.08 g

Therefore, the number of grams of H2SO4 needed to completely react with the given volume of NH3 is approximately 55.08 g.

No, I didn't get that number. If you will post your work I will find the error.

Remember T must be in kelvin. First find n for moles NH3 from PV = nRT, convert that to moles H2SO4, then convert to grams H2SO4.