Astudent burned a 5.00 gram sample of butane. c4h10
find the number of moles of butane in 5.00 grams?
find the number of moles of oxygen that will react with 5.00g of butane?
find the number of grams of oygen that will react with the butane?
find the number of moles of carbon dioxide produced
find the number of grams of carbon dioxide produced
find the number of moles of water produced
find the number of grams of water produced
find the total mass of the reactants and compare to the total mass of the products

a) 0.0860 moles butane

b) 0.559 moles O2

c) 17.9 grams O2

d) 0.344 moles CO2

e) 15.14 grams CO2

f) 0.430 moles H2O

g) 7.75 grams H2O

h) Total mass of reactants = 22.9 grams

Total mass of products ≈ 22.9 grams

The total mass of reactants should be the same as the total mass of products

Explanation:

Step 1: Data given

Mass of butane = 5.00 grams

Molar mass of butane = 58.12 g/mol

Step 2: The balanced equation

2C4H10 + 13O2 → 8CO2 + 10H2O

Step 3: Find the number of moles of butane in 5.00 grams?

Moles butane = mass butane / molar mass butane

Moles butane = 5.00 grams / 58.12 g/mol

Moles butane = 0.0860 moles butane

Step 4: Find the number of moles of oxygen that will react with 5.00g of butane?

For 2 moles butane we need 13 moles O2

For 0.0860 moles butane we'll need 6.5 * 0.0860 = 0.559 moles O2

Step 5: Find the number of grams of oygen that will react with the butane?

Mass O2 = moles O2 * molar mass O2

Mass O2 = 0.559 moles * 32.0 g/mol

Mass O2 = 17.9 grams O2

Step 6: Find the number of moles of carbon dioxide produced

For 2 moles butane consumed, we'll have 8 moles CO2

For 0.0860 moles butane we'll have 4*0.0860 = 0.344 moles CO2

Step 7: Find the number of grams of carbon dioxide produced

Mass CO2 = 0.344 moles * 44.01 g/mol

Mass CO2 = 15.14 grams CO2

Step 8: Find the number of moles of water produced

For 2 moles butane consumed, we'll have 10 moles H2O

For 0.0860 moles we'll have 5*0.0860 = 0.430 moles H2O

Step 9: Find the number of grams of water produced

Mass H2O = 0.430 moles * 18.02 g/mol

Mass H2O = 7.75 grams H2O

Step 10: Find the total mass of the reactants and compare to the total mass of the products

Total mass of reactants = 5.00 grams + 17.9 grams = 22.9 grams

Total mass of products = 15.14 grams + 7.75 grams≈ 22.9 grams

The total mass of reactants should be the same as the total mass of products

hydrocarbons are organic compounds that contain only hydrogens and carbons. they can be saturated (those which contain only carbon-carbon single bonds) or unsaturated (those which contain carbon-carbon double or triple bonds) in nature. hydrocarbons can undergo several reactions like substitution, elimination etc.  

on substituting one or more hydrogen atoms in hydrocarbon it results in the formation of haloalkane. as the halogen atoms are large compared to the carbon and hydrogen atoms, due to which the molecular weight increases and the bond become polar because of presence electronegative halogen atom and results in the increase in boiling point of the haloalkane.

thus, the correct option is (b) that is the boiling point of the new compound increases on substituting a hydrogen atom with a halogen in a hydrocarbon.