How many grams of propane, C3H8, are needed to make 140 grams of water?

85.6 g

Explanation:

Step 1: Write the balanced combustion equation

C₃H₈ + 5 O₂ ⇒ 3 CO₂ + 4 H₂O

Step 2: Calculate the moles corresponding to 140 g of H₂O

The molar mass of H₂O is 18.02 g/mol.

140 g × 1 mol/18.02 g = 7.77 mol

Step 3: Calculate the moles of C₃H₈ needed to produce 7.77 moles of H₂O

The molar ratio of C₃H₈ to H₂O is 1:4. The moles of C₃H₈ needed are 1/4 × 7.77 mol = 1.94 mol.

Step 4: Calculate the mass corresponding to 1.94 moles of C₃H₈

The molar mass of C₃H₈ is 44.10 g/mol.

1.94 mol × 44.10 g/mol = 85.6 g

explanation:

the largest database of organic compounds lists about 10 million substances, which include compounds originating from living organisms and those synthesized by chemists. the number of potential organic compounds has been estimated at 1060—an astronomically high number. the existence of so many organic molecules is a consequence of the ability of carbon atoms to form up to four strong bonds to other carbon atoms, resulting in chains and rings of many different sizes, shapes, and complexities. the simplest organic compounds contain only the elements carbon and hydrogen, and are called hydrocarbons. even though they are composed of only two types of atoms, there is a wide variety of hydrocarbons because they may consist of varying lengths of chains, branched chains, and rings of carbon atoms, or combinations of these structures. in addition, hydrocarbons may differ in the types of carbon-carbon bonds present in their molecules. many hydrocarbons are found in plants, animals, and their fossils; other hydrocarbons have been prepared in the laboratory. we use hydrocarbons every day, mainly as fuels, such as natural gas, acetylene, propane, butane, and the principal components of gasoline, diesel fuel, and heating oil. the familiar plastics polyethylene, polypropylene, and polystyrene are also hydrocarbons. we can distinguish several types of hydrocarbons by differences in the bonding between carbon atoms. this leads to differences in geometries and in the hybridization of the carbon orbitals.