1. Write the Balanced Chemical Equation
First, we need the balanced chemical equation for the reaction between NaClO3 (sodium chlorate) and Cr2O3 (chromium(III) oxide). This reaction is a bit complex and produces several products. It's likely you're interested in the formation of sodium chromate (Na2CrO4). The balanced equation would be:
6 NaClO3 + 2 Cr2O3 → 2 Na2CrO4 + 6 NaCl + 3 O2
2. Convert Masses to Moles
* NaClO3:
* Molar mass of NaClO3 = 106.44 g/mol
* Moles of NaClO3 = (15 g) / (106.44 g/mol) = 0.141 moles
* Cr2O3:
* Molar mass of Cr2O3 = 151.99 g/mol
* Moles of Cr2O3 = (20 g) / (151.99 g/mol) = 0.132 moles
3. Determine the Mole Ratio from the Balanced Equation
The balanced equation shows that 6 moles of NaClO3 react with 2 moles of Cr2O3.
4. Calculate the Limiting Reactant
* NaClO3: If all 0.141 moles of NaClO3 reacted, it would require (0.141 moles NaClO3) * (2 moles Cr2O3 / 6 moles NaClO3) = 0.047 moles of Cr2O3. We have more than enough Cr2O3 (0.132 moles).
* Cr2O3: If all 0.132 moles of Cr2O3 reacted, it would require (0.132 moles Cr2O3) * (6 moles NaClO3 / 2 moles Cr2O3) = 0.396 moles of NaClO3. We don't have enough NaClO3 (only 0.141 moles).
Conclusion:
Since we don't have enough NaClO3 to react with all of the Cr2O3, NaClO3 is the limiting reactant.
