2 M n O 2 + 4 K O H + O 2 + C l 2 → 2 K M n O 4 + 2 K C l + 2 H 2 O , there are 100.0 g of each reactant available. Which reactant is the limiting reagent?

KOH is the limiting reactant

Explanation:

Step 1: Data given

Mass of MnO2 = 100 grams

Mass of KOH = 100 grams

Mass of O2 = 100 grams

Mass of Cl2 = 100 grams

Molar mass of MnO2 = 86.94 g/mol

Molar mass of KOH = 56.11 g/mol

Molar mass of O2 =32.0 g/mol

Molar mass of Cl2 = 70.9 g/mol

Step 2: The balanced equation

2MnO2 + 4KOH + O2 + Cl2 → 2 KMnO4 + 2KCl + 2H2O

Step 3: Calculate moles

Moles = mass / molar mass

Moles MnO2 = 100 grams / 86.94 g/mol

Moles MnO2 =1.15 moles

Moles KOH = 100 grams / 56.11 g/mol

Moles KOH = 1.78 moles

Moles O2 = 100 grams / 32.0 g/mol

Moles O2 = 3.125 moles

Moles Cl2 = 100 grams / 70.9 g/mol

Moles Cl2 = 1.41 moles

Step 4: Calculate the limiting reactant

For 2 moles MnO2 we need 4 moles KOH and 1 mol O2 and 1 mol Cl2 to produce 2 moles KMnO4, 2 moles KCl and 2 moles H2O

KOH is the limiting reactant. It will completely be consumed (1.78 moles). The other reactants are in excess.

There will react:

MnO2: 1.78/ 2 = 0.89 moles

O2: 1.78/4 = 0.445 moles

Cl2: 1.78/4 = 0.445 moles

There wil remain:

MnO2: 1.15 - 0.89 = 0.26 moles

O2: 3.125 - 0.445 = 2.68 moles

Cl2: 1.41 - 0.445 = 0.965 moles

KOH is the limiting reactant

1.solid

2.carbon dioxide

3.helium