Question

What volume of sulfur dioxide gas at 45°C and 723 mmHg will react completely with 1.870 L of oxygen gas at constant temperature and pressure? 2 SO2(g) + O2(g) → 2SO3(g)

Answer 1

Answer:

3.74 L

Explanation:

1.87 L of oxygen gas is used in the reaction

The reaction is

$2SO_2(g)+O_2(g)\rightarrow 2SO_3(g)$

2 moles of $SO_2$ reacts with 1 mole of $O_2$

Same concept can be used for volume

2 L of $SO_2$ reacts with 1 L of $O_2$

Now 1.87 L of $O_2$ is used so

$2\times 1.87\ \text{L}=3.74\ \text{L}$ of $SO_2$ reacts with 1.87 L of $O_2$.

The volume of sulfur dioxide that will react with the required amount of oxygen is 3.74 L.

Answer 2

The volume of sulfur dioxide gas, SO₂ needed to react completely with 1.870 L of oxygen gas at constant temperature and pressure is 3.74 L

We'll begin by calculating the volume of SO₂ that reacted from the balanced equation. This can be obtained as follow:

2SO₂(g) + O₂(g) → 2SO₃(g)

From the balanced equation above,

1 L of O₂ reacted with 2 L of SO₂

• With the above information, we can obtain the volume of SO₂ needed to react completely with 1.870 L of O₂. This can be obtained as illustrated below:

From the balanced equation above,

,1 L of O₂ reacted with 2 L of SO₂

Therefore,

1.870 L of O₂ will react with = 1.87 × 2 = 3.74 L of SO₂

Thus, the volume of SO₂ needed to react completely with 1.870 L of O₂ is 3.74 L

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