This is a limiting reagent (LR) problem. This procedure will find the LR as well as finish the stoichiometry.
1. Write and balance the equation.
2a. Convert 5.07 g butane to mols. mols = grams/molar mass = ?
2b. Do the same and convert g O2 to mols.
3a. Using the coefficients in the balanced equation, convert mols butane to mols product (H2O).
3b. Do the same and convert mols O2 to mols H2O.
3c. It is likely that the two values for mols H2O will not be the same; the correct value in LR problems is ALWAYS the smaller one and the reagent responsible for this is the LR.
4. Using the smaller value of mols H2O, then grams = mols x molar mass = ?
Post your work if you get stuck.
Gaseous butane CH3CH22CH3 will react with gaseous oxygen O2 to produce gaseous carbon dioxide CO2 and gaseous water H2O. Suppose 4.07 g of butane is mixed with 9.8 g of oxygen. Calculate the maximum mass of water that could be produced by the chemical reaction. Round your answer to
2 significant digits.
1 answer
1 answer