Another chemistry question?

Question

im an engine, octane combines with oxygen to form carbon dioxide and water. if 22.8 grams of octane combine completly with 80 grams of oxygen to form 70-4 grams of carbon dioxide , what mass of water if formed??

Answer 1

use the law of conservation of mass.

Mass in = mass out

Mass of octane + Mass of O2 = Mass of CO2 + mass of H2O

22.8+80=70.4+Mass of H2O

Mass of H2O= 32.4

Answer 2

2-C8H18+25-O2 ----> 16-CO2+18-H2O

70.4g(1mol/44g) = 1.6mol(18/16) = 1.8mol H20(18g/1mol) = 32.4g H20

make it a good day

Answer 3

Step 1:  Know use formulæ, rather than names.  Octane = C₈H₁₈.  Carbon dioxide = CO₂.  Water = H₂O.  And of course, oxygen = O₂.

Step 2:  Write the basic chemical equation, as stated above: C₈H₁₈ + O₂ → CO₂ + H₂O

Step 3:  Balance the equation.

Step 4:  Find the molar weight of each component.

Step 5:  Use that the find the number of moles of each component.

Step 6:  Find the limiting factor, if any.  Use that to determine the number of moles of reaction.

Step 7:  Use that to find the number of moles of each product.

Step 8:  Use that to find the product weights.

If you're in Chem 1, things are likely to be simpler.  If you can rely on this, then remember that the goesintas = the comesoutas.  You have 22.8 grams + 80 grams going in, and 70.4 grams + ??? grams coming out.  Use algebra for that:

.....22.8 + 80 = 70.4 + X