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2 years ago

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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 reacta

nt is the limiting reagent?

2 answers:

Sliva [168]2 years ago

6 0

Answer:

The limiting reactant is KOH.

Explanation:

To find the limiting reactant we need to calculate the number of moles of each one:

$\eta = \frac{m}{M}$

<u>Where</u>:

η: is the number of moles

m: is the mass

M: is the molar mass

$\eta_{MnO_{2}} = \frac{100.0 g}{86.9368 g/mol} = 1.15 moles$

$\eta_{KOH} = \frac{100.0 g}{56.1056 g/mol} = 1.78 moles$

$\eta_{O_{2}} = \frac{100.0 g}{31.998 g/mol} = 3.13 moles$

$\eta_{Cl_{2}} = \frac{100.0 g}{70.9 g/mol} = 1.41 moles$

Now, we can find the limiting reactant using the stoichiometric relation between the reactants in the reaction:

$\eta_{MnO_{2}} = \frac{\eta_{MnO_{2}}}{\eta_{KOH}}*\eta_{KOH} = \frac{2}{4}*1.78 moles = 0.89 moles$

We have that between MnO₂ and KOH, the limiting reactant is KOH.

$\eta_{O_{2}} = \frac{\eta_{O_{2}}}{\eta_{Cl_{2}}}*\eta_{Cl_{2}} = \frac{1}{1}*1.41 moles = 1.41 moles$

Similarly, we have that between O₂ and Cl₂, the limiting reactant is Cl₂.

Now, the limiting reactant between KOH and Cl₂ is:

$\eta_{KOH} = \frac{\eta_{KOH}}{\eta_{Cl_{2}}}*\eta_{Cl_{2}} = \frac{4}{1}*1.41 moles = 5.64 moles$

Therefore, the limiting reactant is KOH.

I hope it helps you!

vladimir1956 [14]2 years ago

5 0

Answer:

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

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