Show Limiting reactant calculations based on 3.5 mL acetic acid (using its density), and 2.5 mL...

Show Limiting reactant calculations based on 3.5 mL acetic acid (using its density), and 2.5 mL isopentyl alcohol (using its density) in order to calculate theoretical grams.

Expert Solution

Reaction includes 1 mole each of Acetic acid and isopentyl alcohol

CH3COOH+C5H12O----------------->C7H14O2+H2O

density of Acetic acid=1.05 g/mL

density of isopentyl alcohol=0.81 g/mL

Molecular mass of Acetic acid=60 g

Molecular mass of isopentyl alcohol=88g

given mass =density *volume

Given mass of Acetic acid available=1.05 g/mL*3.5mL=3.675g

Given mass of isopropyl alcohol available=0.81 g/mL*2.5=2.025g

60 g of Acetic acid requires 88g of isopentyl alcohol to react completely.

1g of Acetic acid requires 88/60g of isopentyl alcohol to react completely.

3.675g of Acetic acid requires (88/60)*3.675=5.39g of isopentyl alcohol to react completely.

Here 2.025 g of isopropyl alcohol is available only wwhich is less than required i.e 5.39g to react completely.Hence isopropyl alcohol is limiting reactant.

Please provide rating.

queen_honey_blossom answered 3 years ago

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Show Limiting reactant calculations based on 3.5 mL acetic acid (using its density), and 2.5 mL...

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