For the reaction:
2MnO2+ 4KOH+ O2+ Cl2 yields 2KMnO4+ 2KCL+ 2H20
there is 100. g of each reactant available. Which reagent is the limiting reactant?
convert each 100 g to moles.
Then divide the mole of each by the coefficent for the reactant. The lowest number is your limiting reactant.
To find the limiting reactant, we need to compare the number of moles of each reactant and determine which one is present in the lesser amount. Here's how you can calculate it:
1. Start by writing down the balanced chemical equation for the reaction:
2MnO2 + 4KOH + O2 + Cl2 → 2KMnO4 + 2KCl + 2H2O
2. Convert the given mass of each reactant (100 g) to moles by dividing by their respective molar masses. The molar masses are:
MnO2 = 86.94 g/mol
KOH = 56.11 g/mol
O2 = 32.00 g/mol
Cl2 = 70.90 g/mol
For MnO2:
moles of MnO2 = (100 g MnO2) / (86.94 g/mol MnO2)
For KOH:
moles of KOH = (100 g KOH) / (56.11 g/mol KOH)
For O2:
moles of O2 = (100 g O2) / (32.00 g/mol O2)
For Cl2:
moles of Cl2 = (100 g Cl2) / (70.90 g/mol Cl2)
3. Calculate the stoichiometric ratio between the reactants by dividing the coefficients in the balanced equation. Since MnO2 has a coefficient of 2 in front of it, we divide the number of moles by 2 for both KOH and O2.
Stoichiometric ratio between MnO2 and KOH = moles of MnO2 / (2 * moles of KOH)
Stoichiometric ratio between MnO2 and O2 = moles of MnO2 / (moles of O2)
Stoichiometric ratio between MnO2 and Cl2 = moles of MnO2 / (moles of Cl2)
4. Compare the stoichiometric ratios. The reactant with the lowest ratio will be the limiting reactant because it limits the amount of product formed.
After calculating the stoichiometric ratios for each reactant, you can compare them. The reactant with the lowest ratio will be the limiting reactant.
5. Determine the limiting reactant. The reactant with the smallest stoichiometric ratio is the limiting reactant since it will be completely consumed in the reaction.
Once you find the limiting reactant, you can answer the question of which reagent is the limiting reactant.