CS2(g) + O2(g) CO2(g) + SO2(g)

(a) Consider the unbalanced equation above. What is the maximum mass of SO2 that can be produced when 46.0 g of CS2 and 29.0 g O2 react?
(b) Assume the reaction in part (a) goes to completion. Calculate the mass of excess reactant.

I figured out part a and got an answer of 38.7 but I'm not sure how to do the 2nd part. Please help!

If you did part a correctly, then you know which is the limiting reagent. To do part b, use the limiting reagent and the same kind of stoichiometry you used for part a to determine moles of the other reagent used, then subtract from the initial amount to determine the amount remaining un-reacted. Your answer of 38.7 doesn't help me any. It doesn't have any units (I assume grams) and it doesn't say 38.7 grams of WHAT.

It's 37.9 grams of SO2

To solve part (b), we need to determine which reactant is the excess reactant and calculate the mass of the excess reactant left after the reaction goes to completion.

First, we need to determine the limiting reactant. Let's start by calculating the moles of each reactant using their molar masses.

Molar mass of CS2:
C = 12.01 g/mol
S = 32.07 g/mol
2 x H = 2 x 1.01 g/mol

Molar mass of CS2 = 12.01 + 32.07 + 2.02 = 76.09 g/mol

Moles of CS2 = mass / molar mass = 46.0 g / 76.09 g/mol = 0.604 mol

Molar mass of O2:
O = 16.00 g/mol

Molar mass of O2 = 16.00 g/mol

Moles of O2 = mass / molar mass = 29.0 g / 32.00 g/mol = 0.906 mol

Now, let's look at the balanced equation and the stoichiometry of the reaction:

CS2(g) + O2(g) -> CO2(g) + SO2(g)

From the balanced equation, the stoichiometric ratio between CS2 and SO2 is 1:1. This means that 1 mole of CS2 reacts with 1 mole of SO2.

Since the moles of CS2 is 0.604 mol, the moles of SO2 produced will also be 0.604 mol.

Now, let's compare the moles of O2 and CS2:

Moles of O2 = 0.906 mol
Moles of CS2 = 0.604 mol

The stoichiometric ratio between O2 and SO2 is 1:1. This means that 1 mole of O2 reacts with 1 mole of SO2.

Since the stoichiometric ratio is the same for both reactants, we can conclude that the moles of SO2 produced will be limited by the moles of CS2.

Now, let's calculate the mass of the excess reactant. We know that the moles of SO2 produced is 0.604 mol. We can use this value to determine the amount of O2 required to produce this amount of SO2.

Moles of O2 required = moles of SO2 produced = 0.604 mol

Mass of O2 required = moles of O2 required x molar mass of O2
= 0.604 mol x 32.00 g/mol
= 19.33 g

To find the mass of the excess reactant, subtract the mass of O2 required from the initial mass of O2:

Mass of excess reactant = initial mass of O2 - mass of O2 required
= 29.0 g - 19.33 g
= 9.67 g

Therefore, the mass of the excess reactant is 9.67 g.

To solve part b of the question, we need to determine which reactant is in excess and calculate the mass of that excess reactant.

First, we need to find the limiting reactant. The limiting reactant is the reactant that is completely consumed and determines the maximum amount of product that can be formed. To find the limiting reactant, we can use stoichiometry.

Start by balancing the equation if it is not already balanced:
CS2(g) + 3O2(g) → CO2(g) + 2SO2(g)

Now, we can determine the number of moles for each reactant using the given masses and their respective molar masses:

46.0 g CS2 × (1 mol CS2 / 76.14 g CS2) = 0.604 mol CS2
29.0 g O2 × (1 mol O2 / 32.00 g O2) = 0.906 mol O2

Next, we need to compare the mole ratios between the reactants. From the balanced equation, we can see that the stoichiometric ratio between CS2 and SO2 is 1:2, and between O2 and SO2 is 3:2.

Using these ratios, we can calculate the expected mole amounts of SO2 for both reactants:

For CS2: 0.604 mol CS2 × (2 mol SO2 / 1 mol CS2) = 1.208 mol SO2
For O2: 0.906 mol O2 × (2 mol SO2 / 3 mol O2) = 0.604 mol SO2

Based on the calculations, we can see that the limiting reactant is O2, as it produces the smaller amount of SO2 (0.604 mol).

Now, to calculate the mass of the excess reactant (CS2), we can use stoichiometry again:

0.604 mol O2 × (32.00 g O2 / 1 mol O2) = 19.33 g O2
Since we have 29.0 g of O2, the mass of the excess O2 is:
29.0 g O2 - 19.33 g O2 = 9.67 g O2

So, the mass of the excess reactant (CS2) in part (b) is 9.67 g.