How many grams of water are produced if you combust 1.28 g of C6H14? The balanced equation is: 2C6H14 + 1902 - 12C02 + 14H2O

Explanation:

according to the balance chemical equation

2 moles of c6h14 give us 14 moles of H2O

now mass(in gram) of 2 moles of C6H14 is

moles=mass/molar mass

mass=molar mass*moles

molar mass of C6H14 is 86 therefore

mass=86*2=172

now mass of 14 moles of water is

mass=molar mass*moles

molar mass of water is 18

mass=18*14=252

from the above calculations we conclude that  

172 grams of C6H14 produce 252 grams of water

so 1.28 grams of C6H14 produce =252*1.28/172=1.87 grams of water

result is that 1.28 grams of C6H14 produced 1.87 grams of water

2h2o2--> 2h2o + o2

this reaction is of the spontaneous decomposition of hydrogen peroxide down into water and oxygen.

add 2 molecules of hydrogen peroxide and 2 molecules of water. since oxygen is naturally diatomic, the total number of atoms of each element is now the same on both sides of the equation so it is balanced.

this may not be entirely correct but let me know.