Question

When 7.50 l of sulfur trioxide are produced by the reaction of sulfur dioxide in an excess of oxygen at standard temperature and pressure, how many liters of sulfur dioxide were used? 2so2 (g) + o2 (g) yields 2so3 (g)?

Answer 1

Answer:

$\boxed{\text{7.50 L SO}_{2}}$

Explanation:

For this problem,  we must use Gay-Lussac's Law of  Combining Volumes:

The ratio in which gases react is the same as the ratio of their coefficients in the balanced equation.

The balanced equation is

Theor: 2 L                2 L  

         2SO₂ + O₂ ⟶2 SO₃

V/L:                           7.50

Gay-Lussac's Law tells us that 2 L of SO₂ form 2 L of SO₃.

The volume of SO₂ used was

$V = \text{7.50 L SO}_{3} \times \dfrac{\text{2 L SO}_{2}}{\text{2 L SO}_{3}}\\\\V = \boxed{\textbf{7.50 L SO}_{2}}$

Answer 2

Answer:

7.50

Explanation: