3C3H8 + 20H2CrO4-->9CO2 + 10 Cr2O3 + 32H2O
What mass of chromium (III) oxide will be formed by complete reaction of 1.50g of propane with 35.0ml of 6.26M chromic acid?
This is a limiting reagent problem. I think the easiest way to work them is to calculate two simple stoichiometry problems using each of the reactants.
Here is a complete worked example of a limiting reagent problem. Just follow the steps. For mols H2CrO4 use mols = M x L.
http://www.jiskha.com/science/chemistry/limiting_reagent_problem.html
This helped me with the amount in grams of Cr2O3, but what do I do with the 2nd part, the 35.0 ml of 6.26M chromic acid?
To find the mass of chromium (III) oxide formed, we need to determine the limiting reagent first. The limiting reagent is the reactant that will be completely consumed, thereby determining the maximum amount of product that can be formed.
Step 1: Convert 35.0 mL of 6.26 M chromic acid to moles.
To do this, we need to use the formula:
moles = volume (in L) × molarity
Given:
Volume = 35.0 mL = 0.035 L
Molarity = 6.26 M
moles of chromic acid = 0.035 L × 6.26 M = 0.2195 moles
Step 2: Determine the stoichiometry between chromic acid and chromium (III) oxide.
From the balanced equation:
3C3H8 + 20H2CrO4 → 9CO2 + 10Cr2O3 + 32H2O
The stoichiometric coefficient of chromic acid (H2CrO4) is 20, and the stoichiometric coefficient of chromium (III) oxide (Cr2O3) is 10. This means that 20 moles of chromic acid react to form 10 moles of chromium (III) oxide.
Step 3: Calculate the moles of chromium (III) oxide formed using the stoichiometry.
Using the stoichiometry relationship, we can determine the moles of chromium (III) oxide formed.
moles of Cr2O3 = (moles of chromic acid) × (10 moles of Cr2O3 / 20 moles of H2CrO4)
moles of Cr2O3 = 0.2195 moles × (10 moles / 20 moles)
moles of Cr2O3 = 0.10975 moles
Step 4: Calculate the mass of chromium (III) oxide formed using the molar mass.
To find the mass of chromium (III) oxide formed, we can multiply the moles of Cr2O3 by its molar mass.
Molar mass of Cr2O3 = atomic mass of chromium (Cr) + (3 × atomic mass of oxygen (O))
Molar mass of Cr2O3 = (52.00 g/mol) + (3 × 16.00 g/mol)
Molar mass of Cr2O3 = 152.00 g/mol
mass of Cr2O3 = moles of Cr2O3 × molar mass of Cr2O3
mass of Cr2O3 = 0.10975 moles × 152.00 g/mol
mass of Cr2O3 = 16.68 g
Therefore, 16.68 grams of chromium (III) oxide will be formed by the complete reaction.