Question

For the following reaction, 29.9 grams of sulfur dioxide are allowed to react with 6.26 grams of oxygen gas . sulfur dioxide(g) + oxygen(g) sulfur trioxide(g) What is the maximum mass of sulfur trioxide that can be formed? grams What is the FORMULA for the limiting reagent? What mass of the excess reagent remains after the reaction is complete? grams

Answer 1

The maximum mass of sulfur trioxide formed is 15.55 grams. The formula for the limiting reagent is Oxygen gas (O₂). There is no excess oxygen gas remaining after the reaction is complete.

Step-by-step explanation:

To find the maximum mass of sulfur trioxide formed, we first need to determine the limiting reagent. The limiting reagent is the reactant that is completely consumed in a chemical reaction and determines the amount of product formed. To do this, we compare the moles of each reactant to the balanced equation. Let's start by calculating the moles of sulfur dioxide and oxygen gas.

Mass of sulfur dioxide: 29.9 g
Molar mass of sulfur dioxide: 64.06 g/mol
Moles of sulfur dioxide = Mass / Molar mass = 29.9 g / 64.06 g/mol = 0.467 mol

Mass of oxygen: 6.26 g
Molar mass of oxygen: 32.00 g/mol
Moles of oxygen = Mass / Molar mass = 6.26 g / 32.00 g/mol = 0.195 mol

Using the balanced equation, we can see that the ratio of sulfur dioxide to oxygen gas is 2:1. This means that for every 2 moles of sulfur dioxide, we need 1 mole of oxygen gas. Since we have less oxygen gas (0.195 mol) compared to sulfur dioxide (0.467 mol), oxygen gas is the limiting reagent.

To calculate the maximum mass of sulfur trioxide formed, we need to convert the moles of oxygen gas to moles of sulfur trioxide using the balanced equation. The balanced equation tells us that 2 moles of sulfur dioxide react to form 2 moles of sulfur trioxide. Therefore, 1 mole of oxygen gas reacts to form 1 mole of sulfur trioxide.

Moles of sulfur trioxide = Moles of oxygen gas = 0.195 mol
Mass of sulfur trioxide = Moles * Molar mass = 0.195 mol * 80.06 g/mol = 15.55 g

Therefore, the maximum mass of sulfur trioxide that can be formed is 15.55 grams.

The formula for the limiting reagent is Oxygen gas (O₂), as it is the reactant that is completely consumed in the reaction.

To calculate the mass of the excess reagent remaining after the reaction is complete, we need to determine the amount of the excess reagent that was not consumed. Since oxygen gas is the limiting reagent, all of it is consumed. Therefore, there will be no excess oxygen gas remaining after the reaction is complete.

Answer 2

Answer: a) The maximum mass of sulfur trioxide that can be formed is 31.4 grams

b) The FORMULA for the limiting reagent is
`O_2`

c) Mass of excess reagent remains is 4.8 grams

Step-by-step explanation:

To calculate the moles :

`\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}`

`\text{Moles of} SO_2=(29.9g)/(64g/mol)=0.467moles`

`\text{Moles of} O_2=(6.26g)/(32g/mol)=0.196moles`

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

According to stoichiometry :

1 mole of
`O_2` require = 2 moles of
`SO_2`

Thus 0.196 moles of
`O_2` will require=
`(2)/(1)* 0.196=0.392moles` of
`SO_2`

Thus
`O_2` is the limiting reagent as it limits the formation of product and
`SO_2` is the excess reagent as (0.467-0.392) = 0.075 moles or
`0.075mol* 64g/mol=4.8g` are left.

As 1 mole of
`O_2` give = 2 moles of
`SO_3`

Thus 0.196 moles of
`O_2` give =
`(2)/(1)* 0.196=0.392moles` of
`SO_3`

Mass of
`SO_3=moles* {\text {Molar mass}}=0.392moles* 80g/mol=31.4g`

Thus 31.4 g of
`SO_3` will be produced from the given masses of both reactants.