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

jjj89t'ji; it hink the answer is your mom is not si

when the volume of gas is   changed from   717.73 ml   to 852 ml   the temperature will   change   from 315   c to 452   c

    calculation

this is calculated using the charles law formula

that is,   v1/t1 = v2/t2

          v1=?

        t1 = 315 c   into   kelvin = 315 +273 = 588 k

        v2 = 852   ml

        t2 = 425c + 273 =698 k

make v1   the subject of the formula ( by   multiplying both side by t1)

  v1=( v2 xt1) / t2

v1= (852 ml   x 588 k)/ 698 k= 717.73 ml