Calculate the minimum mass of ammonia needed to produce 396 kg of ammonium sulfate, (NH4)2SO4, and excess sulfuric acid. 2NH3 + H2SO4 → (NH4)2SO4

Mass =  101891.2 g

Explanation:

Given data:

Mass of ammonia needed = ?

Mass of ammonium sulfate produced = 396 Kg = 396 ×1000 = 396000 g

Solution:

Chemical equation:

2NH₃ + H₂SO₄      →      (NH₄)₂SO₄

Number of moles of  (NH₄)₂SO₄:

Number of moles = mass/molar mass

Number of moles = 396000 g / 132.14 g/mol

Number of moles = 2996.8 mol

Now we will compare the moles of ammonia and ammonium sulfate

               (NH₄)₂SO₄          :        NH₃

                      1                   :         2

                  2996.8            :        2/1×2996.8 = 5993.6 mol

Mass of ammonia needed:

Mass = number of moles × molar mass

Mass = 5993.6 mol × 17 g/mol

Mass =  101891.2 g

a. the law of proportion states the combination of 2 atoms in a specific ratio to produce multiple compounds such as the given statement of question ‘1st compound contains 88.8 % cu ’ indicates 88.8 g of cu is present along with 11.2 g oxygen in 100 g of compound whereas the statement ‘2st compound contains 79.9 % cu ’ indicates 79.9 g of cu is present along with 20.1 g oxygen in 100 g of compound

1st compound contains 88.8 g cu

for cu

m = 63 g /mol

atomic % = 88.8 /63 = 1.42

for o

m = 16 g /mol

atomic % = 11.2 /16 = 0.7

b. the emperical formula of 1st compound is

cu1.4o0.7

cu2o

4d^55s^1 ( third choice)