Using the following equation, 2 NH3 (g) + H2SO4 → (NH4)2SO4(aq),Calculate the volume of NH3 (in liters) needed at 28.43ºC and 5.75 atm to reactwith9.78 g of H2SO4.

To solve this problem, we need to follow these steps:

1. First, we need to find the molar mass of H2SO4. The molar mass of H2SO4 is approximately 98.08 g/mol.

2. Next, we calculate the number of moles of H2SO4 in 9.78 g. We use the formula: moles = mass / molar mass. So, moles of H2SO4 = 9.78 g / 98.08 g/mol = 0.0997 mol.

3. From the balanced chemical equation, we can see that the mole ratio of NH3 to H2SO4 is 2:1. Therefore, we need twice the amount of NH3 to react with H2SO4. So, moles of NH3 = 2 * 0.0997 mol = 0.1994 mol.

4. Now, we use the ideal gas law to find the volume of NH3. The ideal gas law is 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 in Kelvin.

5. First, we need to convert the temperature from Celsius to Kelvin. T(K) = 28.43ºC + 273.15 = 301.58 K.

6. The value of R depends on the units used. Since we are looking for the volume in liters and the pressure is given in atm, we use R = 0.0821 L·atm/K·mol.

7. Rearranging the ideal gas law to solve for V, we get V = nRT / P. Substituting the known values, we get V = (0.1994 mol * 0.0821 L·atm/K·mol * 301.58 K) / 5.75 atm = 0.873 L.

So, approximately 0.873 liters of NH3 are needed to react with 9.78 g of H2SO4 under the given conditions.