Answer:
404.8g of (NH4)HSO4 is produced.
Explanation:
Step 1:
Data obtained from the question. This include the following:
Temperature (T) = 10°C = 10°C + 273 = 283K
Pressure (P) = 110KPa = 110/101.325 = 1.09atm
Volume (V) = 75L
Step 2:
Determination of the number of mole of ammonia, NH3.
The number of mole (n) of ammonia, NH3 can be obtained by using the ideal gas equation. This is illustrated below:
Note:
Gas constant (R) = 0.0821atm.L/Kmol
Number of mole (n) =?
PV = nRT
1.09 x 75 = n x 0.0821 x 283
Divide both side by 0.0821 x 283
n = (1.09 x 75) /(0.0821 x 283)
n = 3.52 moles
Step 3:
Determination of the number of mole ammonium hydrogen sulfate produced from the reaction.
We'll begin by writing the balanced equation for the reaction. This is illustrated below:
NH3 + H2SO4 —> (NH4)HSO4
From the balanced equation above,
1 mole of NH3 produced 1 mole of (NH4)HSO4
Therefore, 3.52 moles of NH3 will also produce 3.52 moles of (NH4)HSO4.
Therefore, 3.52 moles of ammonium hydrogen sulfate, (NH4)HSO4 is produced.
Step 4:
Conversion of 3.52 moles of ammonium hydrogen sulfate, (NH4)HSO4 to grams. This is illustrated below:
Molar Mass of (NH4)HSO4 = 14 + (4x1) + 1 + 32 + (16x4) = 115g/mol
Number of mole of (NH4)HSO4 = 3.52 moles
Mass of (NH4)HSO4 =..?
Mass = mole x molar Mass
Mass of (NH4)HSO4 = 3.52 x 115 = 404.8g
Therefore, 404.8g of (NH4)HSO4 is produced.
