Please help with this problem =)

The combustion of liquid chloroethylene, C2H3Cl, yields carbon dioxide, steam, and hydrogen chloride gas.

I. write a balanced equation for the reaction (I sorta get this part)


II. How many moles of oxygen are required to react with 35.00g of chloroethylene?

III. If 25.00g of chloroethylene reacts with with an excess of oxygen, how many grams of each product are formed?

Thanks again =)

I.

2C2H3Cl + 5O2 ==> 4CO2 + 2H2O + 2HCl
Is that what you obtained?

II.
mols in 35.00 g chloroethylene = 35.00/molar mass = 35.00/62.499 = 0.5600 mols. Convert to mols O2.
0.5600 mols C2H3Cl x (5 mol O2/2 mol C2H3Cl) = ??

III.
25.00 g C2H3Cl = ?? mols = 25.00/62.499 = 0.4000 mols C2H3Cl.
a. Convert to mols CO2.
0.4 mol C2H3Cl x (4 mols CO2/2 mol C2H3Cl) = 0.8000 mol CO2.
g CO2 = mols CO2 x molar mass CO2.

b and c are done the same way. Note that you start with what you know (in this case mols C2H3Cl) and multiply by the coefficients in the balanced equation to convert mols C2H3Cl to any of the other materials. How do you know which coefficient is on top and which on bottom? Look at the units. You want the unit of C2H3Cl (in this case) to cancel and you want the unit CO2(in this case) to stay. Likewise, you want the H2O coefficient on top for the water conversion and HCl coefficient on top to convert to HCl.
Post your work if you get stuck.

Hello there! I'm happy to help you with your chemistry problems, and I'll do my best to mix in some humor along the way.

I. Writing a balanced chemical equation can be a bit of a balancing act, but here's the equation for the combustion of chloroethylene:

C2H3Cl + O2 → CO2 + H2O + HCl

Make sure to remember that balancing equations is like juggling. You need to have the same number of each type of atom on both sides, just like keeping all the balls in the air!

II. To determine how many moles of oxygen are needed, we first need to find the molar mass of chloroethylene (C2H3Cl).

C2H3Cl:
C = 2 * 12.01 g/mol = 24.02 g/mol
H = 3 * 1.01 g/mol = 3.03 g/mol
Cl = 1 * 35.45 g/mol = 35.45 g/mol

The total molar mass of chloroethylene is:
24.02 g/mol + 3.03 g/mol + 35.45 g/mol = 62.50 g/mol

Now, let's calculate the number of moles of chloroethylene:
35.00 g / 62.50 g/mol = 0.56 mol

From the balanced equation, we can see that the mole ratio between chloroethylene and oxygen is 1:1. This means that we need 0.56 moles of oxygen to react with 35.00g of chloroethylene.

III. Now, to find the grams of each product formed when 25.00g of chloroethylene reacts with an excess of oxygen, we'll need to go through some mole-to-mole conversions.

First, let's find the number of moles of chloroethylene using the molar mass we calculated earlier:
25.00 g / 62.50 g/mol = 0.40 mol

Since there's an excess of oxygen, it won't limit the reaction. Therefore, we can use the mole ratio from the balanced equation to find the moles of each product formed. From the equation, we can see the mole ratios are:
1 mol C2H3Cl : 1 mol CO2
1 mol C2H3Cl : 1 mol H2O
1 mol C2H3Cl : 1 mol HCl

So, for 0.40 mol of chloroethylene:
CO2: 0.40 mol
H2O: 0.40 mol
HCl: 0.40 mol

To find the grams of each product, we'll need to use their respective molar masses:
CO2: 44.01 g/mol
H2O: 18.02 g/mol
HCl: 36.46 g/mol

Using the mole-to-mass conversion, we can find the grams of each product:
CO2: 0.40 mol * 44.01 g/mol = 17.60 g
H2O: 0.40 mol * 18.02 g/mol = 7.21 g
HCl: 0.40 mol * 36.46 g/mol = 14.58 g

I hope this helps! If you have any more chemistry questions, feel free to ask. Just remember to keep your chemistry jokes balanced - otherwise, they'll be quite... unreactive!

I. To write a balanced equation for the combustion of liquid chloroethylene, we need to determine the reactants and products.

The reactants are liquid chloroethylene, C2H3Cl, and oxygen gas, O2. The products are carbon dioxide gas, CO2, water vapor, H2O, and hydrogen chloride gas, HCl.

The balanced equation is:

C2H3Cl + O2 → CO2 + H2O + HCl

II. To determine how many moles of oxygen are required to react with 35.00g of chloroethylene, we need to use the balanced equation.

We'll use the molar mass of chloroethylene, which is 62.50 g/mol.

First, convert the mass of chloroethylene to moles:

moles of chloroethylene = mass / molar mass
moles of chloroethylene = 35.00g / 62.50 g/mol
moles of chloroethylene ≈ 0.56 mol

From the balanced equation, we can see that the ratio between chloroethylene and oxygen is 1:3. This means that for every 1 mole of chloroethylene, we need 3 moles of oxygen.

So, the number of moles of oxygen required to react with 0.56 mol of chloroethylene is:

moles of oxygen = 3 × moles of chloroethylene
moles of oxygen = 3 × 0.56 mol
moles of oxygen ≈ 1.68 mol

Therefore, 1.68 moles of oxygen are required to react with 35.00g of chloroethylene.

III. To determine the grams of each product formed when 25.00g of chloroethylene reacts with an excess of oxygen, we'll use the balanced equation.

First, we need to find the moles of chloroethylene:

moles of chloroethylene = mass / molar mass
moles of chloroethylene = 25.00g / 62.50 g/mol
moles of chloroethylene ≈ 0.40 mol

From the balanced equation, we can see that the ratio between chloroethylene and carbon dioxide is 1:1, between chloroethylene and water vapor is 1:1, and between chloroethylene and hydrogen chloride is 1:1.

So, the moles of each product formed will be equal to the moles of chloroethylene.

moles of carbon dioxide ≈ 0.40 mol
moles of water vapor ≈ 0.40 mol
moles of hydrogen chloride ≈ 0.40 mol

To find the mass of each product, we'll use the molar masses:

mass of carbon dioxide = moles of carbon dioxide × molar mass of carbon dioxide
mass of water vapor = moles of water vapor × molar mass of water vapor
mass of hydrogen chloride = moles of hydrogen chloride × molar mass of hydrogen chloride

The molar masses are approximately:

molar mass of carbon dioxide ≈ 44.01 g/mol
molar mass of water vapor ≈ 18.02 g/mol
molar mass of hydrogen chloride ≈ 36.46 g/mol

Therefore, the grams of each product formed when 25.00g of chloroethylene reacts with an excess of oxygen are approximately:

mass of carbon dioxide ≈ 0.40 mol × 44.01 g/mol
mass of water vapor ≈ 0.40 mol × 18.02 g/mol
mass of hydrogen chloride ≈ 0.40 mol × 36.46 g/mol

I hope this helps! Let me know if you have any further questions.

Sure! I'd be happy to help you with this problem.

I. To write a balanced equation for the combustion of C2H3Cl, you need to ensure that the number of atoms for each element is equal on both sides of the equation.

The balanced equation for the reaction is:

C2H3Cl + 3 O2 -> 2 CO2 + 2 H2O + 2 HCl

II. To determine the number of moles of oxygen required to react with 35.00g of chloroethylene (C2H3Cl), we can use the stoichiometry of the balanced equation.

First, we need to find the molar mass of C2H3Cl:

C = 12.01 g/mol
H = 1.01 g/mol (2 atoms)
Cl = 35.45 g/mol

Molar Mass of C2H3Cl = 12.01 g/mol + (1.01 g/mol * 3) + 35.45 g/mol = 62.50 g/mol

Next, we can use the molar mass of C2H3Cl to convert grams to moles:

35.00 g C2H3Cl × (1 mol C2H3Cl / 62.50 g C2H3Cl) = 0.560 mol C2H3Cl

From the balanced equation, we know that 1 mol of C2H3Cl reacts with 3 mol of O2. So, the number of moles of O2 required can be calculated as:

0.560 mol C2H3Cl × (3 mol O2 / 1 mol C2H3Cl) = 1.68 mol O2

Therefore, 1.68 moles of oxygen are required to react with 35.00g of chloroethylene.

III. To determine the grams of each product formed when 25.00g of chloroethylene reacts with an excess of oxygen, we can use the stoichiometry of the balanced equation.

First, we need to find the molar mass of C2H3Cl as we calculated before, which is 62.50 g/mol.

We can calculate the moles of C2H3Cl using its molar mass:

25.00 g C2H3Cl × (1 mol C2H3Cl / 62.50 g C2H3Cl) = 0.400 mol C2H3Cl

From the balanced equation, we can see that 1 mol of C2H3Cl reacts to produce 2 mol of CO2 and H2O, as well as 2 mol of HCl. So, the moles of CO2, H2O, and HCl produced will be twice the moles of C2H3Cl:

Moles of CO2 = 0.400 mol C2H3Cl × (2 mol CO2 / 1 mol C2H3Cl) = 0.800 mol CO2
Moles of H2O = 0.400 mol C2H3Cl × (2 mol H2O / 1 mol C2H3Cl) = 0.800 mol H2O
Moles of HCl = 0.400 mol C2H3Cl × (2 mol HCl / 1 mol C2H3Cl) = 0.800 mol HCl

Lastly, we can convert the moles of each product to grams using their respective molar masses:

Mass of CO2 = 0.800 mol CO2 × 44.01 g/mol = 35.21 g CO2
Mass of H2O = 0.800 mol H2O × 18.02 g/mol = 14.42 g H2O
Mass of HCl = 0.800 mol HCl × 36.46 g/mol = 29.17 g HCl

Therefore, when 25.00g of chloroethylene reacts with an excess of oxygen, 35.21g of CO2, 14.42g of H2O, and 29.17g of HCl are formed.

I hope this helps! Let me know if you have any further questions.