balanced equation for the combustion of hydrogen
Best Results From Yahoo Answers Youtube
From Yahoo Answers
Answers:Molar mass H2 = 2g/mol Molar mass O2 = 32g/mol Molar mass H2O = 18g/mol 2H2 + O2 2H2O That is the balanced equation 2mol H2 react with 1 mol O2 to produce 2mol H2O 4g H2 reacts with 32g O2 to produce 36g H2O Your question: 8g H2 will react with 64g O2 to produce 72g H2O But you produced only 55g H2O % yield = 55/72*100 = 76.4% yield
Answers:C25H52 + 38 O2 --> 25 CO2 + 26 H2O
Answers:This is a trick question because all of the equations are balanced combution equations. You don't need to know how to balance an equation to do this one. You need to know how many carbons and hydrogens would be in 2-methyl-1-butene. (Only 1 formula represents 2-methyl-1-butene). Butene has 4 carbons but you also have an additional methyl group so you know the correct answer can only be "c" or "e" (both have 5 carbons). A butene is a butane with a double bond, so it has 2 less carbons that a saturated hydrocarbon. A saturated hydrocarbon has a formula of C(n)H(2n+2) so a hydrocarbon with 1 double bond should have a formula of C(n)H(2n) so "c" is the correct answer.
Answers:Alkane + O2 ---> CO2 + H2O is an unbalanced form. A general balanced form is C(n)H(2n+2) + (1 n+ )O2 ----> nCO2 + (n+1)H2O you might want to double this up to avoid the halves: 2C(n)H(2n+2) + (3n+1)O2 -----> (2n)CO2 + (2n+2)H2O n is the number of carbon atoms in the alkane eg. in ethane n=2: C2H6 + 3 O2 ---> 2CO2 + 3H2O or 2C2H6 + 7O2 ---> 4CO2 + 6H2O Rather than learning the generic formula you can work it out for each case. Learn the first equation I gave you and then try to balance it. Start by using just 1 mole of alkane and then balance the carbons (using CO2), then the hydrogens (using H2O) and then the oxygens (using O2). If you have any fractions then you can multiply the whole equation up to get whole numbers. Edit: The answer by RobA has the general formula of an alkane wrong.