Question

1 lit of oxygen and 3 lit of So, at STP are reacted to produce sulphur trioxide. The the ratio between volume of sulphur trioxide and that of sulphur dioxide after reaction and weight of SO, formed in grams respectively are ​

Answer 1

Answer:

The balanced chemical equation for the reaction between oxygen and sulfur dioxide to produce sulfur trioxide is:

2 SO2(g) + O2(g) → 2 SO3(g)

According to the equation, 2 moles of SO2 react with 1 mole of O2 to produce 2 moles of SO3.

At STP, 1 mole of any gas occupies 22.4 L. Therefore, 1 L of O2 gas contains 1/22.4 = 0.0446 moles of O2 gas. Similarly, 3 L of SO2 gas contains 3/22.4 = 0.134 moles of SO2 gas.

The balanced equation shows that 2 moles of SO2 react with 1 mole of O2 to produce 2 moles of SO3. Therefore, the limiting reactant in this case is O2, as there is only 0.0446 moles of O2 available, which is less than the 0.134 moles of SO2.

Using the mole ratios from the balanced equation, we can calculate the number of moles of SO3 produced:

1 mole of O2 reacts to produce 2 moles of SO3

0.0446 moles of O2 reacts to produce (2/1) x 0.0446 = 0.0892 moles of SO3

Therefore, the volume of SO3 produced at STP can be calculated as:

1 mole of any gas occupies 22.4 L

0.0892 moles of SO3 occupies (22.4 x 0.0892) = 1.998 L

The volume of SO2 produced at STP can be calculated using the mole ratio from the balanced equation:

2 moles of SO2 react to produce 2 moles of SO3

0.134 moles of SO2 reacts to produce (2/2) x 0.134 = 0.134 moles of SO3

1 mole of any gas occupies 22.4 L

0.134 moles of SO2 occupies (22.4 x 0.134) = 3.002 L

Therefore, the ratio between the volume of SO3 and SO2 produced after the reaction is:

Volume of SO3:Volume of SO2 = 1.998 L : 3.002 L

Volume of SO3:Volume of SO2 = 1:1.502