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balanced equation for the combustion of hydrogen
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Question:Write a balanced chemical equation for the combustion of hydrogen gas. If you start with 8 g of hydrogen and you produce 55 g of water what is the % yield?
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: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
Question:Write the balanced equation for the combustion of alkane paraffin (candle wax).
Candle (paraffin wax) = C25 H52 (25 Carbons and 52 Hydrogens)
Answers:C25H52 + 38 O2 > 25 CO2 + 26 H2O
Answers:C25H52 + 38 O2 > 25 CO2 + 26 H2O
Question:What is the balanced chemical equation for the combustion of 2methyl1butene?
a.C6H12(g) + 9 O2(g) 6 CO2(g) + 6 H2O( )
b.C4H8(g) + 6 O2(g) 4 CO2(g) + 4 H2O( )
c.2 C5H10(g) + 15 O2(g) 10 CO2(g) + 10 H2O( )
d.2 C4H10(g) + 13 O2(g) 8 CO2(g) + 10 H2O( )
e.C5H12(g) + 13 O2(g) 5 CO2 (g) + 6 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 2methyl1butene. (Only 1 formula represents 2methyl1butene). 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: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 2methyl1butene. (Only 1 formula represents 2methyl1butene). 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.
Question:
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.
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.
From Youtube
Hydrogen Exploding and Balancing Chemical Equations :When a balloon filled with hydrogen gas is ignited, a synthesis reaction occurs. Diatomic hydrogen and oxygen act as the reactants producing water. You could also call this a combustion reaction since diatomic hydrogen experiences deflagration (rapid burning followed by an explosion) when it mixes with cooler diatomic oxygen. This demonstration is a great way to introduce or revisit the concept of balancing chemical equations. In this demonstration, I introduce the parts of a chemical equation such as: reactants, products, coefficients, subscripts and product sign. In addition, I demonstrate one method of balancing chemical equations that lets you analyze how many atoms of each element exist on either side of the equation. In order for the equation to be balanced, the same number of atoms of each element must be present as products and reactancts. Antoine Lavoisier's work reminds us that matter can not be created or destroyed...it can only change its form (Conservation of Mass). It is pretty cool watching the explosion frame by frame. The entire explosion takes about 4 frames when recorded at 30 frames per second (.13 seconds).
Balancing chemical equations :A sample of how to balance a chemical equation. This is a combustion reaction