Consider the following mixture of SO2(g) and O2(g).If SO2(g) and O2(g) react to form SO3(g),draw a representation of the product mixture assuming the reaction goes to completion. What is the limiting reactant in the reaction? If 96.0 g of SO2 reacts with 32.0 g O2, what mass of product will form?

The reaction is
2 SO2 + O2 -> 2 SO3

32 g of O2 is 1 mole, and that would require 2 moles (128 g) of SO2 to react completely. If you only have 96 g (1.5 moles) of SO2, that is the limiting reactant. 1.5 moles of SO3 would form, and you can figure out the mass from that.

120g

Well, it looks like the SO2 is feeling a bit overwhelmed by the O2 in this reaction. It's like a game of tag where the SO2 is chasing after the O2, but it just can't catch up.

So, let's take a closer look at the game. We start off with 96 grams of SO2 and 32 grams of O2. Now, in order for O2 to react completely, it needs 2 moles of SO2. But here's the trick, we only have 1.5 moles of SO2. So, it looks like the SO2 is the "limiting reactant" in this game of tag.

Since the SO2 is the limiting reactant, it means that it will dictate how much product is formed. So, let's find out how much product we can make from 1.5 moles of SO2.

According to the balanced equation, 2 moles of SO2 react to form 2 moles of SO3. So, for every 1 mole of SO2, we get 1 mole of SO3. Since we have 1.5 moles of SO2, we can expect to get 1.5 moles of SO3 as well.

Now, to find the mass of the product, we need to know the molar mass of SO3. The molar mass of SO3 is approximately 80 grams per mole. So, if we multiply the molar mass by the number of moles, we can find the mass of the product.

1.5 moles of SO3 multiplied by 80 grams per mole gives us a total mass of 120 grams of SO3.

So, in the end, if 96 grams of SO2 reacts with 32 grams of O2, we can expect to get 120 grams of SO3. The SO2 may be the limiting reactant, but at least it can make us laugh with its efforts to catch up to the O2.

To determine the mass of product that will form, we can use the mole ratio from the balanced chemical equation.

From the equation: 2 SO2 + O2 -> 2 SO3

The mole ratio is 2 moles of SO2 : 2 moles of SO3, or 1 mole of SO2 : 1 mole of SO3.

Since we have 1.5 moles of SO2, we can expect to form 1.5 moles of SO3.

To calculate the mass of SO3 formed, we need to multiply the number of moles by the molar mass of SO3:

Molar mass of SO3 = (32.06 g/mol S) + (3 * 16.00 g/mol O) = 80.06 g/mol

Mass of SO3 formed = 1.5 moles * 80.06 g/mol = 120.09 g

Therefore, if 96.0 g of SO2 reacts with 32.0 g of O2, the mass of product (SO3) that will form is 120.09 g.

To draw a representation of the product mixture, first, determine the number of moles of each reactant. This can be done by dividing the given mass of each reactant by their respective molar masses:

Molar mass of SO2 = 64.06 g/mol
Number of moles of SO2 = 96.0 g SO2 / (64.06 g/mol) = 1.5 moles SO2

Molar mass of O2 = 32.00 g/mol
Number of moles of O2 = 32.0 g O2 / (32.00 g/mol) = 1.0 mole O2

Next, determine the stoichiometric ratio between the reactants and the product. The balanced equation tells us that 2 moles of SO2 react with 1 mole of O2 to form 2 moles of SO3.

Given that ratio, for every 2 moles of SO2, we would need 1 mole of O2. However, we only have 1.0 mole of O2, which is less than the stoichiometric amount required. Therefore, O2 is the limiting reactant.

To calculate the mass of the product formed, we need to use the stoichiometry of the reaction. Since 1 mole of SO2 reacts to form 2 moles of SO3, we can calculate the number of moles of SO3 formed:

Number of moles of SO3 = 1.5 moles SO2 x (2 moles SO3 / 2 moles SO2) = 1.5 moles SO3

Finally, we can calculate the mass of the product formed using the molar mass of SO3 (80.06 g/mol):

Mass of SO3 = 1.5 moles SO3 x (80.06 g/mol) = 120.09 g

Therefore, 120.09 grams of SO3 will form when 96.0 grams of SO2 reacts with 32.0 grams of O2.