Given the reaction: SO3 (g) + H2O(g) → H2SO4 which proceeds with a yield of 85.0%calculate the amount of sulfur ic acid that is formed when 120.0kg sulfur trioxide is mixed with 32.4kg water

The answer to your question is 124.95 g of H₂SO₄

Explanation:

Data

yield = 85%

mass of H₂SO₄ = ?

mass of SO₃ = 120 Kg

mass of H₂O = 32.4 kg

Balanced chemical reaction

                       SO₃  +  H₂O  ⇒   H₂SO₄

1.- Calculate the molar mass of the reactants

SO₃ = 32 + (16 x 3) = 32 + 48 = 80 g

H₂O = 2 + 16 = 18 g

2.- Calculate the limiting reactant

theoretical proportion = 80 /18 = 4.44

experimental proportion = 120/32.4 = 3.7

The limiting reactant is SO₃ because the experimental proportion was lower  than the theoretical proportion

3.- Calculate the real amount of SO₃

                    120g 100%

                       x     85%

                       x = (85 x 120)/100

                       x = 102 g of SO₃

4.- Calculate the mass of H₂SO₄

molar mass of H₂SO₄ = 2 + 32 + 64 = 98 g

                 80 g of SO₃ 98 g of H₂SO₄

               102 g of SO₃  x

                 x = (102 x 98)/80

                 x = 124.95 g of H₂SO₄

a pan containing water is placed in a positio n where it is exposed to a 10 mile per hour wind. the pan contains a uniform depth of 1.27 cm of water, is quite wide, and has a length of 2.44 m in the wind direction. the water in the pan is at a constant temperature of 15.5 °c.