Reference no: EM132688692
Consider the following balanced chemical equation. Note: the molar mass of C3822C38g2g2g2l H is 44.10 g/mol, the molar mass of O is 32.00 g/mol, and the molar mass of HO is 18.02 g/mol.
H () + 5 O () à 3 CO () + 4 HO ()
In this reaction 24.0 g of C3H8 is allowed to react with 24.0 g of O2. We ultimately, for this problem, want to know how much grams of water can be formed. However, this can be hard since we have starting amounts of both reactants and we do not know what reactant will run out first (or which reactant is in excess). In order to start this type of problem we will each starting amount of reactant and convert it to grams of product (water in this case). We will break this down into steps below.
a. How much water can be produced from 24.0 g of C3H8? Convert 24.0 g of C3H8 to g of H2O.
b. How much water can be produced from 24.0 g of O2? Convert 24.0 g of O2 to g of H2O.
c. Which reactant made the least amount of product, O2 or C3H8?
d. Based on this, what reactant runs out first O2 or C3H8? (This is called the limiting reactant).
e. Which reactant is in excess and would have some leftover after all of the limiting reactant is used up, O2 or C3H8?