How many grams of potassium permangate are needed to decolorized 1g of 2-pentene in basic solution
To determine the amount of potassium permanganate required to decolorize 1g of 2-pentene in a basic solution, you need to consider the stoichiometry of the reaction between potassium permanganate (KMnO4) and 2-pentene.
1. First, write the balanced chemical equation for the reaction:
KMnO4 + C5H10 → MnO2 + KOH + H2O
According to the equation, one mole of KMnO4 reacts with one mole of 2-pentene to produce one mole of MnO2, one mole of KOH, and one mole of H2O.
2. Find the molar mass of 2-pentene:
The molar mass of C5H10 is:
(5 * atomic mass of carbon) + (10 * atomic mass of hydrogen)
The atomic mass of carbon is approximately 12 g/mol, and the atomic mass of hydrogen is approximately 1 g/mol. Thus,
Molar mass of C5H10 = (5 * 12 g/mol) + (10 * 1 g/mol) = 70 g/mol
3. Convert the mass of 2-pentene to moles:
Moles of 2-pentene = Mass of 2-pentene / Molar mass of C5H10
In this case, since we have 1g of 2-pentene:
Moles of 2-pentene = 1g / 70 g/mol = 0.0143 mol
4. Use stoichiometry to find the moles of KMnO4 needed:
From the balanced equation, we know that 1 mole of KMnO4 reacts with 1 mole of 2-pentene. Therefore, the moles of KMnO4 needed will also be 0.0143 mol.
5. Finally, convert the moles of KMnO4 to grams:
Molar mass of KMnO4 = (atomic mass of potassium) + (atomic mass of manganese) + (4 * atomic mass of oxygen)
The atomic mass of potassium is approximately 39 g/mol, manganese is approximately 55 g/mol, and oxygen is approximately 16 g/mol.
Molar mass of KMnO4 = (39 g/mol) + (55 g/mol) + (4 * 16 g/mol) = 158 g/mol
Mass of KMnO4 = Moles of KMnO4 * Molar mass of KMnO4
Mass of KMnO4 = 0.0143 mol * 158 g/mol = 2.26 g
Therefore, approximately 2.26 grams of potassium permanganate are needed to decolorize 1 gram of 2-pentene in a basic solution.