ethanol, C2H5OH, is responsible for the effts of intoxication felt after drinking alcoholic beverages. When ethanol burns in oxygen, carbon dioxide and water are produced. IF you have 256g of ethanol and 100 grams O2, calculate the amount product produced. You need to write out a balance equation first.

C2H5OH + O2-> CO2 + H2O

Have you tried to balance the equation?

That would be a good place to start.

You will need twice as many CO2's as C2H5OH's. That should be obvious.

You will need 1/3 as many H2O's as C2H5OH's.

So start with

C2H5OH +xO2 -> 2CO2 + 3 H2O

Now all you need to do is balance the O's to figure out what x is.

C2H508+x02_>2C02+3H2=X=23 DUMB***

we start with:

C2H5OH + O2-> CO2 + H2O
Think of this as our skeleton. It cannot be changed. Then, list how many there are of each element. This would be how it would look like:
C: 2 C: 1
H:6 H: 2
O: 3 O: 3
Now, always do hydrogen and oxygen last, so we have to do C first. To make the second carbon have 2, simply put 2 in front of CO2. This changes the O on the right to 5. Now that we have the C's equal, we can balance H. We do this by placing a 3 before the H2O. Now we have this:
C2H5OH + O2-> 2CO2 + 3H2O
C: 2 C: 2
H:6 H: 6
O: 3 O: 7
Now to find the number to put before O2... It is 3...

To balance the equation, we need to make sure that there is an equal number of atoms of each element on both sides of the equation.

First, let's balance the carbon atoms:

C2H5OH + O2 -> 2CO2 + H2O

Next, let's balance the hydrogen atoms:

C2H5OH + O2 -> 2CO2 + 3H2O

Finally, let's balance the oxygen atoms:

C2H5OH + 3O2 -> 2CO2 + 3H2O

Now, we have a balanced equation.

To find the amount of product produced, we need to determine the limiting reactant. The limiting reactant is the reactant that is completely consumed in the reaction and limits the amount of product that can be formed.

To do this, we first calculate the number of moles of each reactant:

Number of moles of ethanol (C2H5OH) = 256g / molar mass of ethanol (46.07 g/mol) = 5.56 moles

Number of moles of oxygen (O2) = 100g / molar mass of oxygen (32.00 g/mol) = 3.125 moles

Now we compare the ratios of reactants to determine the limiting reactant. From the balanced equation, we can see that the ratio of moles of C2H5OH to O2 is 1:3.

Since we have 5.56 moles of C2H5OH and 3.125 moles of O2, we can see that the C2H5OH is the limiting reactant because there is less of it compared to the O2.

Now, using the balanced equation, we can determine the amount of carbon dioxide (CO2) and water (H2O) produced.

From the balanced equation: 1 mole of C2H5OH produces 2 moles of CO2 and 3 moles of H2O.

Therefore, 5.56 moles of C2H5OH will produce (5.56 moles) x (2 moles of CO2 / 1 mole of C2H5OH) = 11.12 moles of CO2

And, 5.56 moles of C2H5OH will produce (5.56 moles) x (3 moles of H2O / 1 mole of C2H5OH) = 16.68 moles of H2O

To calculate the amount of product produced in grams, we multiply the number of moles of each product by their respective molar masses:

Mass of CO2 = (11.12 moles) x (44.01 g/mol) = 488.51 g

Mass of H2O = (16.68 moles) x (18.02 g/mol) = 300.01 g

So, when 256g of ethanol and 100g of oxygen react, the amount of carbon dioxide produced is 488.51 g and the amount of water produced is 300.01 g.